Class 11 - Mathematics
Bonomial Theorem - Exercise 8-Misc
Top Block 1
Question 1:
Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.
Answer:
Given that the first three terms of the expansion are 729, 7290 and 30375, respectively.
Now T1 = nC0 * an-0 * b0 = 729
=> an = 729 …………….1
T2 = nC1 * an-1 * b1 = 7290
=> n * an-1 * b = 7290 …….2
T3 = nC2 * an-2 * b2 = 30375
=> {n(n – 1)/2} * an-2 * b2 = 30375 …….3
Now equation2/equation1
n * an-1 * b/an = 7290/729
=> n * b/a = 10 …….4
Now equation3/equation2
{n(n – 1)/2} * an-2 * b2 /n * an-1 * b = 30375/7290
=> b(n – 1)/2a = 30375/7290
=> b(n – 1)/a = (30375*2)/7290
=> nb/a – b/a = 60750/7290
=> 10 – b/a = 6075/729 (60750 and 7290 is divided by 10)
=> 10 – b/a = 25/3 (6075 and 729 is divided by 243)
=> 10 – 25/3 = b/a
=> (30 – 25)/3 = b/a
=> 5/3 = b/a
=> b/a = 5/3 ……………..5
Put this value in equation 4, we get
n * 5/3 = 10
=> 5n = 30
=> n = 30/5
=> n = 6
Now put this value in equation 1, we get
a6 = 729
=> a6 = 36
=> a = 3
Now from equation 5, we get
b/3 = 5/3
=> b = 5
So, the values of a, b and n are 3, 5 and 6 respectively.
Question 2:
Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.
Answer:
It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by
Tr+1 = nCr an-r br
Assuming that x2 occurs in the (r + 1)th term in the expansion of (3 + ax)9, we obtain
Tr+1 = 9Cr (3)9 – r (ax)r = 9Cr ar xr
Comparing the indices of x in x2 and in Tr + 1, we obtain r = 2
Thus, the coefficient of x2 is
9C2 39-2 a2 = {9!/(2! * 7!)} * 37 a2 = 36 * 37 a2
Assuming that x3 occurs in the (k + 1)th term in the expansion of (3 + ax)9, we obtain
Tk+1 = 9Ck (3)9 – r (ax)k = 9Cr (3)9 – k ak xk
Comparing the indices of x in x3 and in Tk+ 1, we obtain k = 3
Thus, the coefficient of x3 is
9C3 39-3 a3 = {9!/(3! * 6!)} * 36 a3 = 84 * 36 a3
It is given that the coefficients of x2 and x3 are the same.
=> 84 * 36 a3 = 36 * 37 a2
=> 84a = 36 * 3
=> a = 108/84
=> a = 9/7
Thus, the required value of a is 9/7
Question 3:
Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.
Answer:
Using Binomial Theorem, the expressions, (1 + 2x)6 and (1 – x)7, can be expanded as
(1 + 2x)6 = 6C0 + 6C1 (2x) + 6C2 (2x)2 + 6C3 (2x)3 + 6C4 (2x)4 + 6C5 (2x)5 + 6C6 (2x)6
= 1 + 6(2x) + 15(2x)2 + 20(2x)3 + 15(2x)4 + 6(2x)5 + (2x)6
= 1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6
(1 – x)7 = 7C0 – 7C1 (x) + 7C2 (x)2 – 7C3 (x)3 + 7C4 (x)4 – 7C5 (x)5 + 7C6 (x)6 – 7C7 (x)7
= 1 – 7x + 21x2 – 35x3 + 35x4 – 21x5 + 7x6 – x7
Now, (1 + 2x)6 and (1 – x)7
= (1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6) * (1 – 7x + 21x2 – 35x3 + 35x4 – 21x5 + 7x6 – x7)
The complete multiplication of the two brackets is not required to be carried out. Only
those terms, which involve x5, are required.
The terms containing x5 are
1(-21x5) + (-12x)(35x4) + (60x2)(-35x3) + (160x3)(21x2) + (240x4)(-7x) + (192x5)(1) = 171x5
Thus, the coefficient of x5 in the given product is 171
Question 4:
If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer. [Hint: write an = (a – b + b)n and expand]
Answer:
In order to prove that (a – b) is a factor of (an – bn), it has to be proved that
an – bn = k (a – b), where k is some natural number
It can be written that, a = a – b + b
Now, an = (a – b + b)n
= [(a – b) + b]n
= nC0 (a – b)n – nC1 (a – b)n-1 b + nC2 (a – b)n-2 b2 + ………+ nCn-1 (a – b)bn-1 + nCn bn
= (a – b)n – nC1 (a – b)n-1 b + nC2 (a – b)n-2 b2 + ………+ nCn-1 (a – b)bn-1 + bn
=> an – bn = (a – b)n – nC1 (a – b)n-1 b + nC2 (a – b)n-2 b2 + ………+ nCn-1 (a – b)bn-1
=> an – bn = (a – b)[(a – b)n-1 – nC1 (a – b)n-2 b + ………+ nCn-1 bn-1 ]
=> an – bn = k(a – b)
Where k = [(a – b)n-1 – nC1 (a – b)n-2 b + ………+ nCn-1 bn-1 ] is a natural number.
This shows that (a – b) is a factor of (an – bn), where n is a positive integer.
Question 5:
Evaluate. (√3 + √2)6 – (√3 – √2)6
Answer:
Using Binomial Theorem, the expressions, (a + b)6 – (a – b)6 can be expanded as
(a + b)6 = 6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 ab3 + 6C6 b6
= a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6
(a – b)6 = 6C0 a6 – 6C1 a5 b + 6C2 a4 b2 – 6C3 a3 b3 + 6C4 a2 b4 – 6C5 ab3 + 6C6 b6
= a6 – 6a5b + 15a4b2 – 20a3b3 + 15a2b4 – 6ab5 + b6
Now, (a + b)6 – (a – b)6 = [a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6] – [a6 – 6a5b + 15a4b2
– 20a3b3 + 15a2b4 – 6ab5 + b6]
= 2[6a5b + 20a3b3 + 6ab5]
Now, put a = √3 and b = √2, we get
(√3 + √2)6 – (√3 – √2)6 = 2[6(√3)5(√2) + 20(√3)3(√2)3 + 6(√3)(√2)5]
= 2[54√6 + 120√6 + 24√6]
= 2 * 198√6
= 396√6
Question 6:
Find the value of {a2 + √(a2 – 1)}4 + {a2 + √(a2 – 1)}4
Answer :
Firstly, the expression (x + y)4 + (x – y)4 is simplified by using Binomial Theorem.
(x + y)4 = 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 xy3 + 4C4 y4
= x4 + 4x3 y + 6x2y2 + 4xy3 + y4
(x – y)4 = 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 xy3 + 4C4 y4
= x4 – 4x3 y + 6x2y2 – 4xy3 + y4
Now, (x + y)4 + (x – y)4 = [x4 + 4x3 y + 6x2y2 + 4xy3 + y4] – [x4 – 4x3 y + 6x2y2 – 4xy3 + y4]
= 2[x4 + 6x2y2 + y4]
Put x = a2 and y = √(a2 – 1)}, we get
a2 + √(a2 – 1)}4 + {a2 + √(a2 – 1)}4 = 2[(a2)4 + 6(a2)2{(a2 – 1)}2 + {(a2 – 1)}4]
= 2[a8 + 6a4 (a2 – 1) + (a2 – 1)2]
= 2[a8 + 6a6 – 6a4 + a4 + 1 – 2a2]
= 2[a8 + 6a6 – 5a4 – 2a2 + 1]
= 2a8 + 12a6 – 10a4 – 4a2 + 2
Mddle block 1
Question 7:
Find an approximation of (0.99)5 using the first three terms of its expansion.
Answer:
0.99 can be written as,
0.99 = 1 – 0.01
Now, (0.99)5 = (1 – 0.01)5
= 5C0 (1)5 – 5C1 (1)4 (0.01) + 5C2 (1)3(0.01)2
= 1 – 5(0.01) + 10(0.01)2
= 1 – 0.05 + 0.001
= 0.951
Thus, the approximation of (0.99)5 using the first three terms of its expansion is 0.951
Question 8:
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (4√2 + 1/4√3)n is √6 : 1
Answer :
In the expansion, (a + b)n = nC0 an + nC1 an-1 b + nC2 an-2 b2 + …………….. + nCn bn
Fifth term from the beginning = nC4 an-4 b4
Fifth term from the end = nCn-4 a4 bn-4
Therefore, it is evident that in the expansion of (4√2 + 1/4√3)n the fifth term from the beginning
is nC4 (4√2)n-4 (1/4√3)4 and the fifth term from the end is nCn-4 (4√2)4 (1/4√3)n-4
Now, nC4 (4√2)n-4 (1/4√3)4 = nC4 {(4√2)n/(4√2)4}(1/3)
= nC4 {(4√2)n/2}(1/3)
= [n!/{(6 * 4! * (n – 4)!}] * (4√2)n ……………1
and nCn-4 (4√2)4 (1/4√3)n-4 = nCn-4 {2 * (4√3)4/(4√3)n}
= nCn-4 {2 * 3/(4√3)n }
= [6n!/{(4! * (n – 4)!}] * {1/(4√3)n} ……………2
It is given that the ratio of the fifth term from the beginning to the fifth term from the end is
√6 : 1. Therefore, from equation 1 and 2, we get
[n!/{(6 * 4! * (n – 4)!}] * (4√2)n : [6n!/{(4! * (n – 4)!}] * {1/(4√3)n} = √6 : 1
=> [(4√2)n /6] * [(4√3)n/6] = √6
=> (4√6)n /36 = √6
=> (4√6)n = 36√6
=> 6n/4 = 65/2
=> n/4 = 5/2
=> n/2 = 5
=> n = 10
Thus, the value of n is 10
Question 9:
Expand using Binomial Theorem (1 + x/2 – 2/x)4, x ≠ 0
Answer:
Using Binomial Theorem, the given expression (1 + x/2 – 2/x)4 can be expanded as
(1 + x/2 – 2/x)4 = {(1 + x/2) – 2/x}4
= 4C0 (1 + x/2)4 – 4C1 (1 + x/2)3 (2/x) + 4C2 (1 + x/2)2 (2/x)2 – 4C3 (1 + x/2)( 2/x)3 + 4C4 (2/x)4
= (1 + x/2)4 – 4(1 + x/2)3 (2/x) + 6(1 + x/2)2 (4/x2) – 4(1 + x/2)(8/x3) + 16/x4
= (1 + x/2)4 – (1 + x/2)3 (8/x) + (1 + x + x2/4) (24/x2) – (1 + x/2)(32/x3) + 16/x4
= (1 + x/2)4 – (1 + x/2)3 (8/x) + 24/x2 + 24/x + 6 – 32/x3 – 16/x2 + 16/x4
= (1 + x/2)4 – (1 + x/2)3 (8/x) + 8/x2 + 24/x + 6 – 32/x3 + 16/x4
=> (1 + x/2 – 2/x)4 = (1 + x/2)4 – (1 + x/2)3 (8/x) + 8/x2 + 24/x + 6 – 32/x3 + 16/x4 ……….1
Again by using Binomial Theorem, we obtain
(1 + x/2)4 = 4C0 (1)4 + 4C1 (1)3 (x/2) + 4C2 (1)2 (x/2)2 + 4C3 (1)( x/2)3 + 4C4 (x/2)4
= 1 + 4 * (x/2) + 6 * (x2/4) + 4 * (x3/8) + x4/16
=> (1 + x/2)4 = 1 + 2x + 3x2/2 + x3/2 + x4/16 …………..2
(1 + x/2)3 = 3C0 (1)3 + 3C1 (1)2 (x/2) + 3C2 (1)(x/2)2 + 3C3 (x/2)3
=> (1 + x/2)3 = 1 + 3x/2 + 3x2/4 + x3/8 …………..3
From equation 1, 2, and 3, we get
(1 + x/2 – 2/x)4 = 1 + 4 * (x/2) + 6 * (x2/4) + 4 * (x3/8) + x4/16 – [1 + 3x/2 + 3x2/4 + x3/8](8/x) +
8/x2 + 24/x + 6 – 32/x3 + 16/x4
= 1 + 2x + 3x2/2 + x3/2 + x4/16 – 8/x – 12 – 6x – x2 + 8/x2 + 24/x + 6 – 32/x3 + 16/x4
= 16/x + 8/x2 – 32/x3 +16/x4 + x2/2 + x3/2 + x4/16 – 5
Question 10:
Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem.
Answer:
Using Binomial Theorem, the given expression (3x2 – 2ax + 3a2)3 can be expanded as
(3x2 – 2ax + 3a2)3 = [(3x2 – 2ax) + 3a2]3
= 3C0 (3x2 – 2ax)3 + 3C1 (3x2 – 2ax)2 (3a2) + 3C2 (3x2 – 2ax) (3a2)2 + 3C3 (3a2)3
= (3x2 – 2ax)3 + 3(9x2 – 12ax3 + 4a2 x2)(3a2) + 3(3x2 – 2ax) (9a4) + 27a6
= (3x2 – 2ax)3 + 81a2 x4 – 108a3 x3 + 36a4 x2 + 81a4 x2 – 54a5 x + 27a6
=> (3x2 – 2ax + 3a2)3 = (3x2 – 2ax)3 + 81a2 x4 – 108a3 x3 + 117a4 x2 – 54a5 x + 27a6 ………….1
Again by using Binomial Theorem, we obtain
(3x2 – 2ax)3 = 3C0 (3x2)3 – 3C1 (3x2)2 (2ax) + 3C2 (3x2)( 2ax)2 – 3C3 (2ax)3
=> (3x2 – 2ax)3 = 27x6 – 3(9x4)(2ax) + 3(3x2)(4a2 x2) – 8a3 x3
=> (3x2 – 2ax)3 = 27x6 – 54ax5 + 36a2 x4 – 8a3 x3 …………..2
From equation 1 and 2, we get
(3x2 – 2ax + 3a2)3 = 27x6 – 54ax5 + 36a2 x4 – 8a3 x3 + 81a2 x4 – 108a3 x3 + 117a4 x2 – 54a5 x + 27a6
= 27x6 – 54ax5 + 117a2 x4 – 116a3 x3 + 117a4 x2 – 54a5 x + 27a6
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