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BINOMIAL THEOREM
--1
Binomial :- An expression which contains two terms is called a binomial Pascal Triangle:Index
coefficient 1 1
1
1
2
1
2
3
3
1
1
3
1
4
6
4
1
4
5
1
5
10
10
5
1
In each row 1st and last elements ‘1’ remaining terms are obtained by adding the 2 terms Theorem :If n is a positive integer x, a are real number then ( x + a ) n = nc0 x n + nc1 x n −1a + nc2 x n − 2 a 2 + nc3 x n −3 a 3 + ... + ncn a n
Note 1:This theorem is called binomial theorem for positive integer index. Note 2: This expansion contains (n +1) terms Note 3: In the expansion the sum of powers of x and ‘a’ in each term is equal to n. Note 4: In this expansion the power term x is decreased by 1 and the power of a is increased by 1 except in the first term
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www.sakshieducation.com Notes 5: In this expansion (r + 1)th term is called general term denoted by Tr +1 where Tr +1 = ncr . x n −r a r Note 6: In this expansion the coefficients nc0 , nc1 , nc2 ............ncn are called binomial coefficients and these are simply denoted by C0 , C1 , C2 .....Cn n
Note 7: ( x + a ) n = ∑ ncr , x n − r a r r =0
n
Note 8 : ( x − a ) n = ∑ (−1) n ncr . x n − r a r r =0
n
Note 9 : (1 + x) n = ∑ ncr x r r =0
n
Note 10 : (1 − x ) n = ∑ (−1) n ncr x r r =0
Note 11 : Expansion
Nature of n
No. of terms
( x + a) n + ( x − a) n
even
n +1 2
Odd
n +1 2
even
n 2
Odd
n +1 2
( x + a) n − ( x − a) n
Note 12: ( x + a ) n = (a + x) n even though the above expansion are equal Tr +1 in ( x + a) n ≠ Tr +1 in (a + x)n n
x x Note 13 : (a + x) = a 1 + = a n (1 + t ) n where t = a a n
n
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www.sakshieducation.com Note 14: The number of terms in the expansion ( x + a) n + ( x − a) n + ( x + ai ) n + ( x − ai ) n is n + 4 4 where [ . ] is greatest integer function n
Note 15: (a + b + c) n = [a + (b + c)]n = ∑ ncr a n − r (b + c) r r =0
Note 16: Number of terms of (a + b + c) n is (n + 2)k2 Note 17 : Number of terms of ( a1 + a2 + a3 + .... + ar ) n is (n + r − 1)cr −1
np − m b The coefficient of x in ax p + q is ncr a n−r bn −r where r = p+q x n
m
n
np b The constant term in ax p + q is ncr a n−r br where r = p+q x Middle term(s) of ( x + a)n Case (i)
If n is even
Here number of terms = n + 1 ∴n + 1 is odd
Hence only one middle term exist n +1+1 n ∴ middle term = term = + 1 term 2 2 th
Case (ii)
th
If n is odd
Number term n + 1 is even n +1 n + 3 ∴ two middle terms , terms 2 2 th
th
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www.sakshieducation.com Numerically greatest terms :Theorem 1 : If
(n + 1) x x +1
= p a positive integer then pth and ( p + 1)th term are the
numerically greatest terms n the expansion of (1 + x) n ii) If
( n + 1) x = p + F where p is a positive integer and o < F < 1 then ( p + 1)th x +1
term is the numerically greatest term in the expansion of (1 + x) n Note 1 : The numerically greatest term of (a + x)n can be found by writing n
x (a + x) = a 1 + since a n is constant it will not effect the relative numerical b value a of the terms n
n
Note 2 : In the expansion of (1 + x) n if we take x = 1 then
(n + 1) x 1+ x
=
n +1 and the terms 2
are just the binomial coefficient nc0 , nc1 , nc2 ..........ncn thus if n is odd then positive integer and hence Tp =
n +1 =p a 2
ncn −1 nc and Tp +1 = n+1 are the numerically greatest terms 2 2
n +1 1 = p + is not a positive integer and hence Tp +1 = ncn / 2 is the 2 2 numerically greatest coefficient If n is even then
BINOMIAL COEFFICIENTS Theorem : If cr denotes ncr then
i) c0 + c1 + c2 + ..... + cn = 2 n ii) c0 − c1 + c2 − c3 + ...... + (−1) n cn = 0 iii) c0 + c2 + c4 + ...... = c1 + c3 + c5 + ..... = 2n −1
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Note i) :
∑ nc r =0
r
= 2n
n
ii)
∑ (−1) nc r
r
r =0
=0
Theorem 2 : If cr denotes ncr then ac0 + ( a + d ).c1 + ( a + 2 d ).c2 + .... + ( a + nd ).cn = (2a + nd ).2 n −1 Theorem 3: If cr denotes ncr then c1 + 2.c2 x + 3.c3 x 2 + ...... + ncn .x n −1 = n(1 + x ) n −1 Corollary: Prove that c1 + 2.c2 + 3.c3 + .....n.cn = n.2 n −1
We know that c1 + 2.c2 + 3.c3 x 2 + .....n cn x n −1 = n.(1 + x ) n −1
Put x = 1 on both sides c1 + 2.c2 + 3.c3 + .... + n.cn = n.2n −1
Corollary 2 : Prove that c1−1 .c2 + 3.c3 + ...... + n(−1) n −1 cn = 0
We know that (1 + x ) n = c0 + c1 x + c2 x 2 + x3 x 3 + ....... + cn x n n(1 + x ) n −1 = c1 + 2.c2 x + 3.c3 x 2 + .... + n.cn x n −1
Put x = -1 on both sides c1 − 2 c2 + 3 c3 − 4c4 + ..... + ( −1) n −1 n.cn = 0 n
Note i)
∑ r. nc
r
r =0
= n.2n −1
n
ii)
∑ (−1) r. nc r
r =0
r
=0
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www.sakshieducation.com Corollary 3 : Prove that 2.1 c2 + 3.2 c3 x + 4.3 c4 x 2 + .... + n( n − 1)cn x n − 2 = n( n − 1)(1 + x) n − 2
We know that (1 + x ) n = c0 + c1 x + c2 x 2 + c3 x 3 + ..... + cn x n
Different writ x n(1 + x ) n −1 = c1 + 2 c2 x + 3 c3 x 2 + ..... + n cn x n −1
Different again wrt x n(n − 1)(1 + x) n − 2 = 2.1 c2 + 3.2c3 x + 4.3 c4 x 2 + ..... + n(n − 1) cn x n − 2 n
Note :
∑ r (r − 1).nc r =2
r
= n(n − 1).2n − 2
Theorem 4: If cr denotes ncr then c0 +
c (1 + x) n +1 − 1 c1 c x + 2 x 2 + ..... + n x n = 2 3 n +1 (n + 1) x
Theorem 5: If cr denotes ncr then co cr + c1 cr +1 + c2 cr + 2 + ..... + cn − r cr = 2ncn + r
(2n)! (n − r )!(n + r )!
Theorem 6: If cr denotes ncr then c02 + c12 + c22 + ...... + cn2 =
(2n)! (n !) 2
Theorem 7: If cr denotes ncr then prove that c0 c1 + c1c2 + c2 c3 + ..... + cn −1 cn = 2ncn +1 BINOMIAL THEOREM FOR RATIONAL INDEX 1.
If n is a rational number and x < 1 then
1 + nx +
n(n − 1) 2 n(n − 1)(n − 2) 3 x + x + .... = (1 + x)n 2! 3!
2.
(1 + x) −1 = 1 − x + x 2 − x 3 + ..... + x(−1) r x r + .....
3.
(1 + x) −2 = 1 − 2 x + 3x 2 − 4 x 3 + ..... + (−1) r (r + 1) x r + .....
4.
(1 + x) −3 = 1 − 3x + 6 x 2 − 10 x3 + ..... + (−1)r
(r + 1)(r + 2) r x + ..... 2
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5.
(1 + x )
−
p q
2
3
x p( p + q) x p ( p + q )( p + 2q ) x = 1− p + + + ...∞ 1.2 q 1.2.3 q q
n(n + 1) 2 n(n + 1)(n + 2) 3 x + x + ..... 1.2 1.2.3 n(n + 1)(n + 2)....(n + r − 1) r x + ......∞ 1.2.3......r
6. (1 + x)− n = 1 + nx +
7.
(1 − x) −1 = 1 + x + x 2 + x 3 + .... + x r + .....∞
8.
(1 − x) −2 = 1 + 2 x + 3 x 2 + 4 x 3 + .... + (r + 1) x r + .....∞
9.
(1 − x) −3 = 1 + 3x + 6 x 2 + 10 x3 + .... +
(r + 1)(r + 2) r x + .....∞ 2 2
− p/q
x p( p + q) x = 1+ p + + ......∞ 1.2 q q
10.
(1 − x )
11.
If n is a positive integer then (1 + x) − n = 1 − nc1 x + (n + 1)c2 x 2 − (n + 2)c3 x 3 + ......∞ (1 + x) − n = 1 + nc1 x + (n + 1)c2 x 2 + (n + 2)c3 x 3 + ......∞
General term (1 + x)− n = (−1) r (n + r − 1)cr x r General term (1 − x)− n = (n + r − 1)cr x r
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www.sakshieducation.com EXERCISE – 5(a) I. 1.
Expand the following using binomial theorem. (i) (4x + 5y)
7 2 (ii) x + y 4 3
7
2p 3p (iii) − 7 5 i) (4x + 5y)7
5
6
(iv) (3 + x – x2)4
Sol. ( 4x + 5y ) = 7
7
C0 (4x)7 (5y)0 + 7 C1 (4x)6 (5y)1 + 7 C2 (4x)5 (5y)2 + 7 C3 (4x)4 (5y)3 + 7 C4 (4x)3 (5y) 4 + 7 C5 (4x) 2 (5y)5 +
7
C6 (4x)1 (5y)6 + 7 C7 (4x)0 + (5y)7 7
=
∑ 7Cr (4x)7−r (5y)r r =0
7 2 ii) x + y 4 3
5
5
7 2 Sol. x + y = 4 3 5
5
4
1
5
4
2 7 7 C 4 x y + 5 C5 y 3 4 4 5
=
3
2
2
3
2 2 7 2 7 2 7 C0 x + 5C1 x y +5 C2 x y + 5C3 x y + 3 3 4 3 4 3 4
∑ r =0
5
2 Cr x 3
2p 3p iii) − 7 5
5− r
7 y 4
5
r
6
6
6
5
1
2p 3p 2p 2p 3q Sol. − = 6C0 − 6 C1 7 5 5 5 7 4
2
3
3
2
4
1
5
2p 3q 2p 3q + 6 C 2 − 6 C3 5 7 5 7
2p 3q 2p 3q 3q + 6 C 4 − 6 C5 + 6 C 6 5 7 5 7 7 6
2p = ∑ (−1) Cr 5 r =0 r 6
6− r
3q 7
5
r
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www.sakshieducation.com iv) (3 + x – x2)4 Sol. (3 + x − x 2 ) 4 = [(3 + x) − x 2 ]4
= 4 C0 (3 + x) 4 − 4 C1 (3 + x)3 x 2 + 4C2 (3 + x) 2 (x 2 ) 2 − 4C3 (3 + x)1 (x 2 )3 + 4C4 (x 2 ) 4 = (3 + x) 4 − 4(3 + x)3 x 2 + 6(3 + x) 2 x 4 − 4(3 + x)x 6 + x 8 = [ 4 C0 (3) 4 + 4 C1 (3)3 x + 4 C2 (3) 2 x 2 + 4 C3 (3)x 3 + 4C4 x 4 ] − 4[3 C0 (3)3 + 3
C1 (3) 2 x + 3C2 (3)x 2 + 3C3 x 3 ]x 2 + 6[ 2 C6 (3) 2 + 2 C1 (3)x + 2 C2 x 2 ]x 4 − 4(3 + x)x 6 + x 8
= 81 + 108x + 54x 2 + 12x 3 + x 4 −4x 2 (27 + 27x + 9x 2 + x 3 ) + 6x 4 (9 + 6x + x 2 ) − 4x 6 (3 + x) + x 8 = 81 + 108x + (54 − 108)x 2 + (12 − 108)x 3 + (1 − 36 + 54)x 4 + (−4 + 36)x 5 + (6 − 12)x 6 + (−4)x 7 + x 8
2.
= 81 + 108x − 54x 2 − 96x 3 + 19x 4 + 32x 5 − 6x 6 − 4x 7 + x 8 Write down and simplify 2x 3y + i) 6th term in 2 3 th ii) 7 term in (3x – 4y)10
9
14
3p iii) 10 term in − 5q 4 th
8
3a 5b iv) rth term in + (1 ≤ r ≤ 9) 5 7 i)
2x 3y 6th term in + 2 3
2x 3y + Sol. 6 term in 2 3
9
9
th
9
2x 3y The general term in + is 2 3 2x Tr +1 = 9 Cr 3 Put r = 5 4
9− r
3y 2 5
r
4
5
2x 3y 2 3 T6 = C5 = 9C5 x 4 y5 3 2 3 x 9 × 8 × 7 × 6 (24 ) 35 4 5 = x y = 189x 4 y5 1× 2 × 3 × 4 34 25 ii) 7th term in (3x – 4y)10 Sol. General term in (3x – 4y)10 is Tr +1 = (−1) r 10 Cr (3x)10−r (4y) r 9
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Put r = 6 T7 = (−1)6 10 C6 (3x) 4 (4y)6 = 10 C4 (3) 4 (4)6 x 4 y6 10 × 9 × 8 × 7 4 6 4 6 3 (4) x y = 280(12)5 x 4 y6 = 1× 2 × 3 × 4 14
3p iii) 10th term in − 5q 4
14
3p Sol. General term in − 5q 4 14− r
3p Tr +1 = Cr 4 14
is
(−5q) r 14− r
3p = (−1) r 14 Cr (5q) r 4 Put r = 9 5 3p 9 14 T10 = (−1) C9 (5q)9 4 5
3 = − 14 C5 (5)9 p5q 9 4 14 ×13 ×12 ×11×10 35 9 5 9 5 pq =− 1 × 2 × 3 × 4 × 5 45
−(2002)35 ⋅ 59 5 9 pq = 45 8 3a 5b iv) rth term in + (1 ≤ r ≤ 9) 5 7 8
3a 5b Sol. The general term in + is 5 7 8− r
r
3a 5b Tr +1 = Cr 5 7 Replace r by r – 1, we get 8− r +1 r −1 3a 5b 8 Tr = C(r −1) 5 7 8
3a = C(r −1) 5 8
9− r
5b 7
r −1
; 1≤ r ≤ 9
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3.
Find the number of terms in the expansion of 9
3a b (i) + 4 2 (iii) (2x + 3y + z)7
(ii) (3p + 4q)14
9
3a b i) + 4 2 Sol. Number of terms in (x + a)n is (n + 1), where n is a positive integer. 9
3a b Hence number of terms in + are :9 + 1 = 10 4 2 14 ii) (3p + 4q) Sol. Number of terms in (3p + 4q)14 are :14 + 1 = 15 iii) (2x + 3y + z)7
(n + 1)(n + 2) , where n is a positive integer. 2 (7 + 1)(7 + 2) 8 × 9 = = 36 Hence number of terms in (2x + 3y + z)7 are : 2 2
Sol. Number of terms in (a + b + c)n are
II. 1. Find the coefficient of i) x
−6
10
4 in 3x − x
13
11
ii) x
3 in 2x 2 + 3 x
2 iii) x 2 in 7x 3 − 2 x iv) x
i)
x
−6
−7
9
2x 2 5 − 5 in 4x 3
7
10
4 in 3x − x
10
4 Sol. The general term in 3x − x
is
r
4 Tr +1 = (−1) r 10 Cr (3x)10−r x r 10 10− r r 10 − r − r = (−1) Cr 3 (4) x
= (−1) r 10Cr 310−r (4) r x10−2r
...(1)
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For coefficient of x–6 , put 10 – 2r = –6 ⇒ 2r = 10+6 = 16 ⇒ r = 8 Put r = 8 in (1) T8+1 = (−1)8 10 C8 310−8 (4)8 x10−16 = 10 C8 32 48 x −6 10
4 ∴ Coefficient of x in 3x − x 10 2 8 10 2 8 C8 3 4 = C2 3 4 –6
is :
10 × 9 × 9 × 48 = 405 × 48 1× 2 13 2 3 in 2x + 3 x =
ii)
x11
13
3 Sol. The general term in 2x 2 + 3 x 2 13− r
Tr +1 = Cr (2x ) 13
3 3 x
is :
r
= 13Cr (2)13−r 3r x 26−2r x −3r = 13Cr (2)13−r (3) r x 26−5r ...(1) For coefficient of x11, put 26 – 5r = 11 ⇒ 5r = 15 ⇒ r = 3 Put r = 3 in (1) T3+1 = 13C3 (2)10 (3)3 x 26−15 T4 =
13 ×12 ×11 10 3 11 ⋅ 2 ⋅3 ⋅ x 1× 2 × 3 13
3 ∴ Coefficient of x11 in 2x 2 + 3 x 10 3 (286)(2 )(3 ) 2 iii) x in 7x 3 − 2 x
is :
9
2
9
2 Sol. The general term in 7x 3 − 2 is x 3 9− r
Tr +1 = (−1) Cr (7x ) r 9
2 2 x
r
= (−1) r 9 Cr (7)9−r (2) r x 27 −3r x −2r ...(1) = (−1) r 9 Cr (7)9−r (2) r x 27 −5r 2 For coefficient of x , put 27 – 5r = 2 ⇒ 5r = 25 ⇒ r = 5 www.sakshieducation.com
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Put r = 5 in (1) T5+1 = (−1)5 9 C5 (7) 4 (2)5 x 27 −25
9×8× 7 × 6 4 5 2 = − 9 C 4 7 4 ⋅ 25 ⋅ x 2 = − 7 ⋅2 x 1× 2 × 3 × 4 9
2 ∴ Coefficient of x2 in 7x 3 − 2 is :–126 × 74 × 25. x
iv) x
−7
2x 2 5 − 5 in 4x 3
7
7
2x 2 5 Sol. The general term in − 5 is 4x 3
2x 2 Tr +1 = (−1) ⋅ Cr 3
7−r
5 5 4x
r 7
2 = (−1) ⋅ Cr 3
7−r
r 7
r
r
5 14−2r −5r x x 4
7−r
r
2 5 ∴ Tr +1 = (−1) r 7 Cr x14−7r ...(1) 3 4 For coefficient of x–7 , put 14 – 7r = –7 ⇒ 7r = 21 ⇒ r = 3 Put r = 3 in equation (1) 4 3 2 5 T3+1 = (−1)3 7 C3 x14−21 3 4 4
=
3
−7 × 6 × 5 2 5 −7 x 1× 2 × 3 3 4 7
2x 2 5 ∴ Coefficient of x in − 5 is : 4x 3 –7
= −35 × 2.
1 53 −4375 ⋅ = 324 34 22
Find the term independent of x in the expansion of 10
x1/ 2 4 (i) − 2 3 x
3 (ii) 3 + 5 x x
25
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www.sakshieducation.com 2x 2 15 (iv) + 4x 5
14
7 (iii) 4x 3 + 2 x
9
10
i)
x1/ 2 4 − 2 x 3
10
x1/ 2 4 Sol. The general term in − 2 x 3 10− r
Tr+1 = (−1)
r 10
= (−1)
r 10
= (−1)
r 10
x1/ 2 Cr 3
10− r
1 Cr 3
4 2 x
r
5−
r 2
(4) ⋅ x r
10− r
1 Cr 3
is
(4)
r
⋅ x −2r
r 5− − 2r 2 ⋅x 10−5r
10− r
1 = (−1) Cr (4) r ⋅ x 2 3 For the term independent of x, 10 − 5r Put = 0 ⇒ 5r = 10 ⇒ r = 2 2 Put r = 2 in eq.(1) 8 1 2 0 2 10 T2+1 = (−1) C2 4 ⋅ x 3 r 10
T3 =
ii)
... (1)
80 729
3 3 +5 x x
25
3 Sol. The general term in 3 + 5 x x 3 Tr +1 = Cr 3 x 25
25
is
25− r
(5 x ) r
= 25Cr (3) 25−r (5) r ⋅ x −1/ 3(25−r) x r / 2 = 25Cr (3)25−r (5) r ⋅ x
−
25 r r + + 3 3 2
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25− r
−
50 + 2r +3r 6
= Cr (3) (5) ⋅ x ...(1) For term independent of x, put −50 + 5r = 0 ⇒ 5r = 50 ⇒ r = 10 6 Put r = 10 in equation (1), T10+1 = 25C10 (3)15 (5)10 x 0 25
r
i.e. T11 = 25C10 (3)15 (5)10
14
7 iii) 4x 3 + 2 x
14
Sol.
7 The general term in 4x 3 + 2 x 7 Tr +1 = 14 Cr (4x 3 )14−r 2 x
is
r
= 14 Cr (4)14−r (7) r x 42−3r x −2r = 14 Cr (4)14−r (7) r x 42−5r ...(1) For term independent of x, Put 4x – 5r = 0 ⇒ r = 42/5 which is not an integer. Hence term independent of x in the given expansion is zero.
iv)
2x 2 15 + 4x 5
9
9
2x 2 15 Sol. The general term in + is 5 4x
2x 2 Tr +1 = Cr 5 9
2 = Cr 5
9−r
9
9−r
9−r
15 4x
r
r
15 18−2r − r ⋅x x 4 r
2 15 = 9 Cr x18−3r ... (1) 5 4 For term independent of x, put 18 – 3r = 0 ⇒ r = 6 Put r = 6 in equation (1)
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6
23 36 × 56 2 15 T6+1 = 9 C6 x 0 = 9 C6 ⋅ 3 ⋅ 5 46 5 4 =
3.
9 × 8 × 7 36 × 56 37 × 53 × 7 ⋅ = 1× 2 × 3 46 27
Find the middle term(s) in the expansion of 10
11
3x (i) − 2y 7
3 (ii) 4a + b 2
20
3 (iv) 3 + 5a 4 (iii) (4x + 5x ) a n Sol. The middle term in (x + a) when n is even is T n +1 , when n is odd, we have two 2
3 17
2
middle terms, i.e. T n +1 and T n +3 . 2
2
10
3x − 2y 7 Sol. n = 10 is even, we have only one middle term. 10 i.e. + 1 = 6th term 2 i)
10
3x ∴ T6 in − 2y 7
is :
5
35 3x = 10C5 (−2y)5 = −(10 C5 ) 5 ⋅ 25 (xy)5 7 7 5
6 = − C5 x 5 y 5 7 10
11
ii)
3 4a + b 2
Sol. Here n = 11 is an odd integer, we have two middle terms, i.e.
= 7th and 7th terms are middle terms. 11
3 T6 in 4a + b 2
is :
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n +1 n +3 and terms 2 2
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35 3 = 11C5 (4a)6 b = 11C5 (4)6 5 a 6 b5 2 2 =
11×10 × 9 × 8 × 7 7 5 6 5 2 ⋅3 ⋅a b 1× 2 × 3 × 4 × 5
= 77 × 28 × 36 × a 6 b5 11
3 T7 in 4a + b 2
is : 6
36 3 = 11C6 (4a)5 b = 11C5 (4)5 6 a 5 b6 2 2 =
11×10 × 9 × 8 × 7 4 6 5 6 2 ⋅3 ⋅a b 1× 2 × 3 × 4 × 5
= 77 × 25 × 37 × a 5 b6 iii) (4x 2 + 5x 3 )17 Sol.
(4x 2 + 5x 3 )17 = [x 2 (4 + 5x)]17 = x 34 (4 + 5x)17 ...(1) 17 Consider (4 + 5x) ∵ n = 17 is odd positive integer, we have two middle terms. They are 9th and 10th terms are middle terms. T9 in (4 + 5x)17 is = 17 C8 (4)17−8 (5x)8 = 17 C8 (4)9 58 x 8 ∴ T9 in (4x2 + 5x3)17 is = x 34 17 C8 ⋅ 49 ⋅ 58 ⋅ x 8 = 17 C8 49 ⋅ 58 ⋅ x 42
T10 in (4 + 5x)17 is = 17 C9 417 −9 ⋅ (5x)9 = 17 C9 ⋅ 48 ⋅ 59 ⋅ x 9 ∴ T10 in (4x 2 + 5x 3 )17 is
= x 34 ⋅ 17 C9 48 ⋅ 59 ⋅ x 9 = 17 C9 ⋅ 48 ⋅ 59 ⋅ x 43 3 iv) 3 + 5a 4 a
20
20 + 1 Sol. Here n = 20 is even positive integer, we have only one middle term, i.e. = 2 11th term
3 T11 in 3 + 5a 4 a
20
is
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3 = 20C10 3 a = 20C10 4.
i)
20−10
(5a 4 )10
310 10 40 20 ⋅ 5 ⋅ a = C10 (15)10 ⋅ a10 30 a
Fin the numerically greatest term (s) in the expansion of 7 i) (4 + 3x)15 when x = 2 1 4 ii) (3x + 5y)12 when x = and y = 2 3 13 iii) (4a – 6b) when a = 3, b = 5 4 iv) (3 + 7x)n when x = , n = 15 5 7 (4 + 3x)15 when x = 2 15
3 Sol. Write (4 + 3x) = 4 1 + x 4 15
15
3 = 415 1 + x 4
…(1) 15
3 First we find the numerically greatest term in the expansion of 1 + x 4 3 (n + 1) | x | Write X = x and calculate 4 1+ | x | 3 3 7 21 Here | X |= X = × = 4 4 2 8 (n + 1) | x | 15 + 1 21 Now = ⋅ 21 8 1+ | x | 1+ 8 16 × 21 336 17 = = = 11 29 29 29 17 Its integral part m = 11 = 11 29 15
3 Tm+1 is the numerically greatest term in the expansion 1 + x 4 4
11
3 3 7 Tm +1 = T12 = C11 x = 15C11 ⋅ 4 4 2 ∴ Numerically greatest term in (4 + 3x)15 15
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and
www.sakshieducation.com 11 11 21 15 (21) = 4 C11 = C4 3 2 8 15 15
ii)
1 4 and y = 2 3
(3x + 5y)12 when x =
12
5y Sol. Write (3x + 5y) = 3x 1 + 3x 12
12
12 12
5 y = 3 x 1 + 3x 12
5 y n On comparing 1 + with (1 + x) , we get 3 x 5 y 5 (4 / 3) 5 8 40 n = 17, x = ⋅ = = ⋅ = 3 x 3 (1/ 2) 3 3 9 40 (12 + 1) (n + 1) | x | 9 Now = 40 1+ | x | 1+ 9 13 × 40 520 30 = = = 10 49 49 49 which is not an integer. 30 ∴ m = 10 = 10 49 12
5y N.G. term in 1 + 3x
is 10
10
5 (4 / 3) 5 y 12 Tm +1 = T11 = C10 = C10 × 3 x 3 (1/ 2) 12
10
10
5 8 40 = C10 × = 12 C10 3 3 9 12 ∴ N.G. term in (3x + 5y) is 12
12 12 1 12
=3 2
10
40 C10 9
2
= 12C10
10
312 (22 )10 × (10)10 12 3 20 = C10 12 2 10 2 (3 ) 2 3
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www.sakshieducation.com iii) (4a – 6b)13 when a = 3, b = 5 13
6b Sol. Write (4a – 6b) = 4a 1 − 4a 13
13
3 b = (4a)13 1 − 2a 13
3 b n On comparing 1 − with (1 + x) 2a −3 b We get n = 13, x = 2 a −3 5 −5 x= × = 2 3 2 −5 5 (13 + 1) 14 × (n + 1) | x | 2 2 = = Now 5 −5 1+ | x | 1+ 1+ 2 2 70 = = 10 which is an integer. 7 Hence we have two numerically greatest terms namely T10 and T11. 13
3 b 3 b 13 T10 in 1 − = C9 − ⋅ 2 a 2a
9
9
3 5 5 = 13C9 ⋅ = 13C9 2 3 2 13 T10 in (4a – 6b) is
9
9
5 5 = (4a) ⋅ C9 = (4 × 3)13 ⋅ 13C9 2 2
9
13 13
9
5 = C9 (12) (12) = 13C9 (12) 4 (30)9 2 13
4
9
13
3 b T11 in 1 − 2a
10
−3 b is = 13C10 ⋅ 2 a 10
10
3 5 5 = C10 × = 13C10 2 3 2 13 ∴ N.G. term in (4a – 6b) is 13
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10
5 5 = (4a)13 ⋅ 13C10 = (4 × 3)13 ⋅ 13C10 2 2
= (12)13 ⋅ 13C10
10 510 13 3 10 5 = C (12) ⋅ (12) ⋅ 10 210 210
= 13C10 (12)3 (30)10 4 , n = 15 5
iv) (3 + 7x)n when x = Sol.
7 Write (3 + 7x)n = 3 1 + x 3 7 = 3 1 + x 3
n
n
n
7 Now we first find N.G. term in 1 + x 3 On comparing with (1 + x)n, we get 7 7 4 28 X= x= × = ⇒ n = 15 3 3 5 15 28 (15 + 1) (n + 1) | x | 15 Now = 28 1+ | x | 1+ 15 16 × 28 448 18 = = = 10 43 43 43 Its integral part (m) = 10 ∴ Tm+1 = T11 is the N.G. term 15
7 T11 in 1 + x = 3
n
10
15
7 C10 x 3
10
10
7 4 28 = C10 × = 15C10 3 5 15 n ∴N.G. term in (3 + 7x) is 15
10
10
28 28 = 315 ⋅ 15C10 = 15C10 ⋅ 35 15 5 5. i) Sol.
Prove the following 2 ⋅ C0 + 5 ⋅ C1 + 8 ⋅ C2 + ... + (3n + 2) ⋅ Cn
= (3n + 4) ⋅ 2n −1 Let S = 2 ⋅ C0 + 5 ⋅ C1 + 8 ⋅ C2 + ...
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... + (3n − 1) ⋅ Cn −1 + (3n + 2)Cn ∵ Cn = C0 , Cn −1 = C1... S = (3n + 2)C0 + (3n − 1)C1 + (3n − 4)C2 + ... .... + 5Cn −1 + 2 ⋅ Cn Add ———————————————— 2S = (3n + 4)C0 + (3n + 4)C1 + (3n + 4)C2 + ... + (3n + 4)Cn
= (3n + 4)(C0 + C1 + C2 + ... + C n ) = (3n + 4)2n ∴ S = (3n + 4) ⋅ 2n −1 ii) C0 − 4 ⋅ C1 + 7 ⋅ C2 − 10 ⋅ C3 + .... = 0 Sol. 1, 4, 7, 10 … are in A.P. Tn+1 = a + nd = 1 + n(3) = 3n + 1 ∴ C0 − 4 ⋅ C1 + 7 ⋅ C2 − 10 ⋅ C3 + ...(n + 1)terms
= C0 − 4 ⋅ C1 + 7 ⋅ C2 − 10 ⋅ C3 + .... + (−1) n (3n + 1)Cn n
= ∑ (−1) r (3r + 1)Cr r =0 n
{
= ∑ (−1) r (3r)Cr + (−1) r Cr r =0
n
n
r =0
r =0
}
= 3 ⋅ ∑ (−1) r r ⋅ Cr + ∑ (−1) r ⋅ Cr = 3(0) + 0 = 0 ∴ C0 − 4 ⋅ C1 + 7 ⋅ C2 − 10 ⋅ C3 + ... = 0 iii)
C1 C3 C5 C7 2n − 1 + + + + ... = 2 4 6 8 n +1
Sol.
C1 C3 C5 C7 + + + + ......... 2 4 6 8 n C1 n C3 n C5 n C7 = + + + + ... 2 4 6 8 n n(n − 1)(n − 2) n(n − 1)(n − 2)(n − 3)(n − 4) = + + + ... 2 4 × 3! 6 × 5! 1 (n + 1)n (n + 1)n(n − 1)(n − 2) (n + 1)n(n − 1)(n − 2)(n − 3)(n − 4) = + + .... + ... n + 1 2! 4! 6! 1 (n +1) = C2 + (n +1) C4 + (n +1) C6 + ... n +1
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1 2n − 1 1 (n +1) n (n +1) (n +1) (n +1) = 2 − 1 = C + C + C + ... − C 0 2 4 0 n +1 n +1 n +1 C C C C 2n − 1 ∴ 1 + 3 + 5 + 7 + ... = 2 4 6 8 n +1 3 9 27 3n 4n +1 − 1 iv) C0 + C1 + C2 + C3 + ... + Cn = 2 3 4 n +1 3(n + 1) Sol. Let S = 3 32 33 3n C0 + C1 + C2 + C3 + ... + Cn …(1) 2 3 4 n +1 ⇒3S= 32 33 34 3n +1 C0 ⋅ 3 + C1 + C2 + C3 + ... + Cn ...(2) 2 3 4 n +1 ⇒ (n + 1)3 ⋅ S =
= (n + 1)C0 ⋅ 3 + (n + 1)C1 ⋅ ⇒ (n + 1)3 ⋅ S
32 33 34 3n +1 + (n + 1)C2 ⋅ + (n + 1)C3 ⋅ + ... + (n + 1)Cn ⋅ 2 3 3 n +1
= (n +1) C1 ⋅ 3 + (n +1) C2 ⋅ 32 + (n +1) C3 ⋅ 33 + ...... + (n +1) Cn +1 ⋅ 3n +1 = (1 + 3) n +1 − (n +1) C0 = 4n +1 − 1 4n +1 − 1 ∴S = 3(n + 1) v)
C0 + 2 ⋅ C1 + 4 ⋅ C2 + 8 ⋅ C3 + ... + 2n ⋅ Cn = 3n
Sol. L.H.S. = C0 + 2 ⋅ C1 + 4 ⋅ C2 + 8 ⋅ C3 + ... + 2n ⋅ Cn
= C0 + C1 (2) + C2 (22 ) + C3 (23 ) + ... + Cn (2n ) = (1 + 2)n = 3n Note : (1 + x) n = C0 + C1 ⋅ x + C2 x 2 + ... + Cn x n 6. Using binomial theorem, prove that 50n – 49n – 1 is divisible by 492 for all positive integers n. Sol. 50n – 49n – 1 = (49 + 1)n – 49n – 1 = [ n C0 (49) n + n C1 (49) n −1 + n C2 (49) n −2 + ... + n Cn −2 (49) 2 + n Cn −1 (49) + n Cn (1)] − 49n − 1
= (49) n + n C1 (49)n −1 + n C2 (49)n −2 + ... + n Cn −2 (49)2 + (n)(49) + 1 − 49n − 1 = 492 [(49)n −2 + n C1 (49)n −3 + n C2 (49)n −4 + ... + ..... + ..... + n Cn −2 ] = 492 [a positive integer] Hence 50n – 49n – 1 is divisible by 492 for all positive integers of n.
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Using binomial theorem, prove that 54n + 52n – 1 is divisible by 676 for all positive integers n. Sol. 54n + 52n – 1 = (52)2n + 52n – 1 = (25) 2n + 52n − 1
= (26 − 1) 2n + 52n − 1 = [ 2n C0 (26) 2n − 2n C1 (26)2n −1 + 2n C2 (26) 2n −2 − ..... + 2n C2n −2 (26) 2 − 2n C2n −1 (26) + = 2n C0 (26) 2n − 2n C1 (26)2n −1 + 2n C2 (26)2n −2 − ..... + 2n C2n −2 − 2n(26) + 1 + 52n − 1 = (26) 2 [ 2n C0 (26) 2n −2 − 2n C1 (26) 2n −3 + 2n C2 (26) 2n −4 + ... + 2n C2n −2 ] is divisible by (26)2 = 676 ∴54n + 52n – 1 is divisible by 676, for all positive integers n. 8.
If (1 + x + x 2 ) n = a 0 + a1x + a 2 x 2 + ... + a 2n x 2n , then prove that
i)
a 0 + a1 + a 2 + ... + a 2n = 3n
3n + 1 2 3n − 1 iii) a1 + a 3 + a 5 + ... + a 2n −1 = 2 n −1 iv) a 0 + a 3 + a 6 + a 9 + ... = 3 ii)
a 0 + a 2 + a 4 + ... + a 2n =
Sol. (1 + x + x 2 ) n = a 0 + a1x + a 2 x 2 + ... + a 2n x 2n Put x = 1, ∴ a0 + a1 + a2 + … + a2n = (1+1+1)n = 3n …(1) Put x = –1, a0 – a1 + a2 – … + a2n = (1–1+1)n = 1 …(2) i) a 0 + a1 + a 2 + ... + a 2n = 3n
ii) (1) + (2) ⇒ 2( a 0 + a 2 + a 4 + ... + a 2n ) = 3n + 1 3n + 1 2 iii) (1) – (2) ⇒ 2(a1 + a 3 + a 5 + ... + a 2n −1 ) = 3n − 1 ∴ a 0 + a 2 + a 4 + ... + a 2n =
∴ a1 + a 3 + a 5 + ... + a 2n −1 = iv) Put x = 1 a 0 + a1 + a 2 + ... + a 2n = 3n
3n − 1 2 …(a)
Hint : 1 + ω + ω = 0 ; ω = 1 Put x = ω a 0 + a1ω + a 2 ω2 + a 3ω3 + ... + a 2n ω2n = 0 …(b) 2
3
Put x = ω2
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2n
C2n (1)] + 52n − 1
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a 0 + a1ω2 + a 2ω4 + a 3ω6 + ... + a 2n ω4n = 0 …(c) Adding (a), (b), (c) 3a 0 + a1 (1 + ω + ω2 ) + a 2 (1 + ω2 + ω4 ) + a 3 (1 + ω3 + ω6 ) + ... + a 2n (1 + ω2n + ω4n ) = 3n
⇒ 3a 0 + a1 (0) + a 2 (0) + 3a 3 + ... + ... = 3n 3n = 3n −1 3 9. If the coefficients of (2r + 4)th term and (3r + 4)th term in the expansion of (1 + x)21 are equal, find r. Sol. T2r+4 in (1 + x)21 is = 21C2r+3(x)2r+3 …(1) 21 21 3r+3 ...(2) T3r+4 in (1 + x) is = C3r+3(x) ⇒ Coefficients are equal ⇒ 21C2r+3 = 21C3r+3 ⇒ 21 = (2r + 3) + (3r + 3) (or) 2r + 3 = 3r + 3 ⇒ 5r = 15 ⇒ r = 3 (or) r = 0 Hence r = 0, 3. III. 1. If 36, 84, 126 are three successive binomial coefficients in the expansion of (1 + x)n, find n. Sol. Let nCr–1, nCr, nCr+1 are three successive binomial coefficients in the expansion of (1 + x)n, find n. Then nCr–1 = 36, nCr = 84 and nCr+1 = 126 n C 84 n − r +1 7 ⇒ = Now n r = r 3 Cr −1 36 3n − 3r + 3 = 7r ⇒ 3n = 10r − 3 ∴ a 0 + a 3 + a 6 + a 9 + ... =
3n + 3 ...(1) =r 10 n C 126 n−r 3 ⇒ n r +1 = ⇒ = 84 r +1 2 Cr ⇒
⇒ 2n − 2r = 3r + 3 ...(2) ⇒ 2n = 5r + 3 3n + 3 ⇒ 2n = 5 + 3 from (1) 10 3n + 3 + 6 ⇒ 2n = ⇒ 4n = 3n + 9 ⇒ n = 9 2 2.If the 2nd, 3rd and 4th terms in the expansion of (a + x)n are respectively 240, 720, 1080, find a, x, n. Sol. T2 = 240 ⇒ nC1 an–1 x = 240 …(1) n n–2 2 T3 = 720 ⇒ C2 a x = 720 ...(2) n n–3 3 T4 = 1080 ⇒ C3 a x = 1080 ...(3) www.sakshieducation.com
www.sakshieducation.com n (2) C a n −2 x 2 720 ⇒ n 2 n −1 = (1) 240 C1a x
n −1 x = 3 ⇒ (n − 1)x = 6a ...(4) 2 a n C a n −3 x 3 1080 (3) ⇒ n 3 n −2 2 = (2) 720 C2a x
⇒
n−2 x 3 = 3 a 2 ...(5) ⇒ 2(n − 2)x = 9a (4) (n − 1)x 6a n −1 2 ⇒ = ⇒ = (5) 2(n − 2)x 9a 2n − 4 3 ⇒ 3n − 3 = 4n − 8 ⇒ n = 5 From (4), (5 – 1)x = 6a ⇒ 4x = 6a 3 ⇒x= a 2 3 Substitute x = a , n = 5 in (1) 2 3 3 5 C1 ⋅ a 4 ⋅ a = 240 ⇒ 5 × a 5 = 240 2 2 ⇒
480 = 32 = 25 15 3 3 ∴ a = 2, x = a = (2) = 3 2 2 ∴ a = 2, x = 3, n = 5 a5 =
If the coefficients of rth, (r+1)th and (r+2)th terms in the expansion of (1 + x)th are in A.P. then show that n2 – (4r + 1)n + 4r2 – 2 = 0. Sol. Coefficient of Tr = nCr–1 Coefficient of Tr+1 = nCr Coefficient of Tr+2 = nCr+1 Given nCr–1, nCr, nCr+1 are in A.P. ⇒ 2 nCr = nCr–1 + nCr+1 3.
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⇒2
n! n! n! = + (n − r)!r! (n − r + 1)!(r − 1)! (n − r − 1)!(r + 1)!
⇒
2 1 1 = + (n − r)r (n − r + 1)(n − r) (r + 1)r
⇒
1 2 1 1 − = n − r r n − r + 1 (r + 1)r
1 2n − 2r + 2 − r 1 = n − r r(n − r + 1) r(r + 1) ⇒ (2n − 3r + 2)(r + 1) = (n − r)(n − r + 1)
⇒
⇒ 2nr + 2n − 3r 2 − 3r + 2r + 2 = n 2 − 2nr + r 2 + n − r ⇒ n 2 − 4nr + 4r 2 − n − 2 = 0 ∴ n 2 − (4r + 1)n + 4r 2 − 2 = 0 14
4.
32
Find the sum of the coefficients of x and x
–18
3 in the expansion of 2x 3 − 2 . x
14
3 Sol. The general term in 2x 3 − 2 x 3 Tr +1 = 14 Cr (2x 3 )14−r − 2 x
is :
r
= (−1) r 14 Cr (2)14−r ⋅ (3) r ⋅ x 42−r ⋅ x −2r = (−1) r ⋅ 14 Cr 214−r (3) r x 42−5r ...(1) 32 From coefficients of x , Put 42 – 5r = 32 ⇒ 5r = 10 ⇒ r = 2 Put r = 2 in equation (1) T3 = (−1) 2 14 C2 (2)12 (3) 2 ⋅ x 42−10 = 14C2 (2)12 (3) 2 ⋅ x 32 Coefficient of x32 is 14C2 (2)12 (3) 2 …(2) For coefficient of x–18 Put 42 – 5r = –18 ⇒ 5r = 60 ⇒ r = 12 Put r = 12 in equation (1) T13 = (−1)12 14C12 (2) 2 (3)12 ⋅ x 42−60 = 14C12 (2) 2 (3)12 ⋅ x −18 ∴ Coefficient of x–18 is 14C12(2)2 312 Hence sum of the coefficients of x32 and x–18 is
14
C2 (2)12 (3)2 + 14 C12 (2)2 (3)12 .
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5.
If P and Q are the sum of odd terms and the sum of even terms respectively in the expansion of (x + a)n then prove that (i) P2 – Q2 = (x2 – a2)n (ii) 4PQ = (x + a)2n – (x – a)2n Sol. (x + a) n = n C0 x n + n C1x n −1a + n C2 x n −2a 2 + n
C3 x n −3a 3 + ... + n Cn −1xa n −1 + n Cn a n
= ( n C0 x n + n C2 x n −2a 2 + n C4 x n −4a 4 + ...) + ( n C1x n −1a + n C3 x n −3a 3 + n C5 x n −5a 5 + ...) = P+Q (x − a) n = n C0 x n − n C1x n −1a + n C2 x n −2a 2 − n C3 x n −3a 3 + ... + n Cn (−1) n a n = ( n C0 x n + n C2 x n −2 a 2 + n C4 x n −4 a 4 + ...) − ( n C2 x n −1a + n C3 x n −3a 3 + n C5 x n −5a 5 + ...) = P−Q i) P2 – Q2 = (P + Q)(P − Q) = (x + a)n (x – a)n = [(x + a) (x – a)]n = (x2 – a2)n ii) 4PQ = (P + Q)2 – (P – Q)2 = [(x + a)n]2 – [(x – a)n]2 = (x + a)2n – (x – a)2n If the coefficients of 4 consecutive terms in the expansion of (1 + x)n are a1, a2, a3, a4 respectively, then show that a3 a1 2a 2 + = a1 + a 2 a 3 + a 4 a 2 + a 3 Sol. Given a1, a2, a3, a4 are the coefficients of 4 consecutive terms in (1 + x)n respectively. Let a1 = nCr–1, a2 = nCr, a3 = nCr+1, a4 = nCr+2 a3 a1 L.H.S. : + a1 + a 2 a 3 + a 4 a3 a1 = + a a a1 1 + 2 a 3 1 + 4 a1 a3 1 1 = + n n C C 1 + n r 1 + n r+2 Cr −1 Cr +1 6.
=
=
1 1 + n − r +1 n − r −1 1+ 1+ r r+2 r r+2 r + r + 2 2(r + 1) + = = n +1 r + 2 + n − r −1 n +1 n +1
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R.H.S :
2a 2
a a 2 1 + 3 a2 2 2(r + 1) = = = L.H.S n−r n 1 + 1+ r +1
2 C 1 + n r +1 Cr a3 a1 2a 2 + = ∴ a1 + a 2 a 3 + a 4 a 2 + a 3 Prove that ( 2n C0 ) 2 − ( 2n C1 ) 2 + ( 2n C2 ) 2 − ( 2n C3 ) 2 + ... + ( 2n C2n ) 2 = (−1) n n
7.
2a 2 = a 2 + a3
Sol. (x + 1) 2n = 2n C0 x 2n + 2n C1x 2n −1 +
2n
2n
Cn
C2 x 2n −2 + ... + 2n C2n ...(1)
(x − 1) 2n = 2n C0 − 2n C1x + 2n C2 x 2 + ... + 2n C2n x 2n Multiplying eq. (1) and (2), we get ( 2n C0 x 2n + 2n C1x 2n −1 + 2n C2 x 2n −2 + ... + 2n C2n )
...(2)
( 2n C0 − 2n C1x + 2n C2 x 2 + ... + 2n C2n x 2n ) = (x + 1) 2n (1 − x) 2n = [(1 + x)(1 − x)]2n = (1 − x 2 ) 2n 2n
= ∑ 2n Cr (− x 2 ) r r =0
Equating the coefficients of x2n ( 2n C0 ) 2 − ( 2n C1 ) 2 + ( 2n C2 ) 2 − ( 2n C3 ) 2 + .... + ( 2n C2n ) 2 = (−1)n 8.
2n
Cn
Prove that (C0 + C1 )(C1 + C2 )(C2 + C3 )...(Cn −1 + Cn ) =
Sol. (C0 + C1 )(C1 + C2 )(C2 + C3 )...(Cn −1 + Cn ) =
(n + 1) n ⋅ C0 ⋅ C1 ⋅ C2 ⋅ ...Cn n!
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www.sakshieducation.com C C C = C0 1 + 1 ⋅ C1 1 + 2 ...Cn −1 1 + n C1 Cn −1 C0 n n n C C C = 1 + n 1 1 + n 2 ...... 1 + n n C0 C1C2 ...Cn −1 C0 C1 Cn −1
n n −1 1 = 1 + 1 + ... 1 + 2 n 1 Cn ⋅ C1 ⋅ C2 ⋅ ...Cn −1[C0 = Cn ]
1 + n 1 + n 1+ n = ...... C1 ⋅ C2 ⋅ ...Cn −1 ⋅ Cn 1 2 n =
(1 + n) n C1C2 ...Cn n!
∴ (C0 + C1 )(C1 + C2 )(C2 + C3 )...(Cn −1 + Cn ) =
(n + 1) n ⋅ C0 ⋅ C1 ⋅ C2 ⋅ ...Cn n! n
9.
1 Find the term independent of x in (1 + 3x) 1 + . 3x n
n
1 3x + 1 Sol. (1 + 3x) n 1 + = (1 + 3x) n 3x 3x
n
n
1 2n 2n 1 2n = (1 + 3x) = n n ∑ ( Cr )(3x)r 3 ⋅ x r =0 3x The term independent of x in n
1 1 (1 + 3x) 1 + is n ( 2n Cn )3n = 2n Cn 3 3x 2 10 10. If (1 + 3x − 2x ) = a 0 + a1x + a 2 x 2 + ... +a 20 x 20 then prove that n
i) a 0 + a1 + a 2 + ... + a 20 = 210 ii) a 0 − a1 + a 2 − a 3 + ... + a 20 = 410 Sol. (1 + 3x − 2x 2 )10 = a 0 + a1x + a 2 x 2 + ... +a 20 x 20 i) Put x = 1 (1 + 3 − 2)10 = a 0 + a1 + a 2 + ... + a 20 ∴ a 0 + a1 + a 2 + ... + a 20 = 210
ii) Put x = –1 (1 − 3 − 2)10 = a 0 − a1 + a 2 + ... + a 20 ∴ a 0 − a1 + a 2 − a 3 + ... + a 20 = (−4)10 = 410
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www.sakshieducation.com 11. If R, n are positive integers, n is odd, 0 < F < 1 and if (5 5 + 11) n = R + F , then prove that i) R is an even integer and ii) (R + F)F = 4n. Sol. i) Since R, n are positive integers, 0 < F < 1 and (5 5 + 11) n = R + F
Let (5 5 − 11) n = f Now, 11 < 5 5 < 12 ⇒ 0 < 5 5 − 11 < 1 ⇒ 0 < (5 5 − 11) n < 1 ⇒ 0 < f < 1 ⇒ 0 > −f > −1 ∴ –1 < –f < 0 R + F – f = (5 5 + 11) n − (5 5 − 11) n n C0 (5 5) n + n C1 (5 5) n −1 (11) + = n C2 (5 5) n −2 (11) 2 + ... + n Cn (11) n n C0 (5 5) n − n C1 (5 5) n −1 (11) + − n C2 (5 5) n −2 (11) 2 + ... + n Cn (−11) n = 2 n C1 (5 5) n −1 (11) + n C3 (5 5) n −3 (11) 2 + ... = 2k where k is an integer. ∴ R + F – f is an even integer. ⇒ F – f is an integer since R is an integer. But 0 < F < 1 and –1 < –f < 0 ⇒ –1 < F – f < 1 ∴F–f=0⇒F=f ∴ R is an even integer. ii) (R + F)F = (R + F)f, ∵ F = f = (5 5 + 11) n (5 5 − 11) n n
= (5 5 + 11)(5 5 − 11) = (125 − 121) n = 4n
∴ (R + F)F = 4n.
12. If I, n are positive integers, 0 < f < 1 and if (7 + 4 3) n = I + f , then show that (i) I is an odd integer and (ii) (I + f)(I – f) = 1. Sol. Given I, n are positive integers and (7 + 4 3) n = I + f , 0 < f < 1
Let 7 − 4 3 = F Now 6 < 4 3 < 7 ⇒ −6 > −4 3 > −7 ⇒ 1 > 7 − 4 3 > 0 ⇒ 0 < ( 7 − 4 3 )n < 1 ∴0