Interactive ISC Class 12 Maths Solutions OP Malhotra Chapter 13 Binomial Theorem Ex 13(a) engage students in active learning and exploration.
S Chand Class 11 ICSE Maths Solutions Chapter 13 Binomial Theorem Ex 13(a)
Question 1.
What is the number of terms in the expansion of each of the following?
(i) [(x -2y)3]3
(ii) (5a + 7b)8
(iii) \(\left(6 x-\frac{1}{x^3}\right)^{17}\)
(iv) (4x2 + 12xy + 9y2)9
(v) (3 + 2√5)18 – (3 – 2√5)18
(vi) (5 + 7x)15 + (5 – 7x)15
(vii) (a + bx)17 – (a – bx)17
Solution:
(i) [(x – 2y)3]3 = (x – 2y)9
Thus number of terms in the expansion of [(x – 2y)3]3 be (9 + 1) i.e. 10.
(ii) The no. of terms in the expansion of (5a + 7b)8 be (8 + 1) i.e. 9
(iii) The number of terms in the expansion of \(\left(6 x-\frac{1}{x^3}\right)^{17}\) is (17 + 1) i.e. 18.
(iv) Now (4x2 + 12xy + 9y2)9 = [(2x + 3y)2]9 = (2x + 3y)18
Thus, the no. of terms in given expansion be 18 + 1 i.e. 19.
(v) We know that, the no. of terms in the expansion of (a + b)n – (a – b)n be \(\frac { n }{ 2 }\) if n be even.
The number of terms in the expansion of (3 + 2√5)18 – (3 – 2√5)18 be \(\frac { 18 }{ 2 }\) i.e. 9 [Here n = 18]
(vi) We know that the no. of terms in the expansion of (a + b)n + (a – b)n be \(\frac { n + 1 }{ 2 }\) if n is odd.
Thus the no. of terms in the expansion of (5 + 7x)15 + (5 – 7x)15 be \(\frac { 15 + 1 }{ 2 }\) i.e. 8 .
(vii) We know that the number of terms in the expansion of (a + b)n – (a – b)n be \(\frac { n + 1 }{ 2 }\) if n be odd. Hence, the no. of terms in the expansion of (a + bx)17 – (a – bx)17 be \(\frac { 17 + 1 }{ 2 }\) i.e. 9 .
Question 2.
Write out the expansions of the following:
(a) (3x – y)4
(b) (3 + 2x2)4
(c) \(\left(x-\frac{y}{2}\right)^4\)
(d) \(\left(2 x+\frac{y}{2}\right)^5\)
(e) (1 + 2x)7
(f) \(\left(\frac{2}{x}-\frac{x}{2}\right)^5\), x ≠ 0
Solution:
We know that binomial theorem for positive integral index be given by
Question 3.
Using binomial theorem, expand [(x + y)5 + (x – y)5] and hence find the value of [(√3 + 1)5 – (√3 – 1)5].
Solution:
By binomial theorem, we have
(x + y)5 = 5C0 x5 y0 + 5C1 x4 y1 + 5C2 x3y2 + 5C3x2y3 + 5C4 xy4 + 5C5 x0y5
= x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5 …(1)
Similarly (x – y)5 = x5 – 5x4y + 10x3 y2 – 10x2 y3 + 5x y4 – y5 …(2)
On adding eqn. (1) and eqn. (2); we have
(x + y)5 + (x – y)5 = 2x5 + 20x3 y2+10xy4 …(3)
putting x =√3 and y = 1 in eqn. (3); we have
(√3 + 1)5 + (√3 – 1)5 = 2(√3)5 + 20(√3)3(1)2 + 10 × √3(1)4
= 18√3 + 60√3 + 10√3 = 88√3
Subtracting eqn. (2) from eqn. (1); we have
(x + y)5 – (x – y)5 = 10x4y + 20x2y3 + 2y5 …(4)
putting x = √3 and y = 1 in eqn. (4); we have
(√3 + 1)5 – (√3 – 1)5 = 10(√3)4 × 1 + 20(√3)2 × 13 + 2 × 15 = 10 × 9 + 20 × 3 + 2 = 152
Question 4.
Expand (2 + x)5 – (2 – x)5 in ascending powers of x and simplify your result.
Solution:
Now (2 + x)5 = 5C0 25 x0 + 5C1 24 x1 + 5C2 23 x2 + 5C3 23 x2 + 5C3 22x3 + 5C4 2x4 + 5C5 20 x5
=32 + 5 × 16x + \(\frac{5 \times 4}{2}\) × 8x2 + 10 × 4x3 + 5 × 2 × x4 + x5
=32 + 80x +80x2 + 40x3 + 10x4 + x5 …(1)
Similarly (2 – x)5 = 32 – 80x + 80x2 – 40x3 + 10x4 – x5 …(2)
eqn. (1) – eqn. (2) gives ; (2 + x)5 – (2 – x)5 = 160x + 80x3 + 2x5
Question 5.
Evaluate the following :
(i) (2 + √5)5 + (2 – √5)5
(ii) (√3 + 1)5 – (√3 – 1)5
Hence, show in (ii), without using tables, that the value of (√3 + 1)5 lies between 152 and 153.
Solution:
We know that (x + a)n = nCo xn ao + nC1 xn – 1 a1 + …. + nCn x0 an
(i) (2 + √5)5 = 5C0 25 (√5)0 + 5C1 24 √5 + 5C2 23 (√5)2 + 5C3 22 (√5)3 + 5C4 21 (√5)4 + 5C5 20 (√5)5 …(1) and (2 – √5)5 = 5C0 25 (- √5)0 + 5C1 24 (- √5) + 5C2 23 (- √5)2 + 5C3 22 (- √5)3 + 5C4 21 (- √5)4 + 5C5 20 (- √5)5
= 5C0 25 (√5)0 – 5C1 24 √5 + 5C2 23 (√5)2 – 5C3 22 (√5)3 + 5C4 2 (√5)4 – 5C5 (√5)5 …(2)
on adding eqn. (1) and eqn. (2); we have
(2 + √5)5 + (2 – √5)5 = 2 C0 25 (√5)0 + 2 × 5C2 23 (√5)2 + 2 × 5C4 × 2 (√5)4
= 2 × 1 × 32 + \(\frac{2 \times 5 \times 4}{2}\) × 8 × 5 + 2 × 5 × 2 × 25 = 64 + 800 + 500 = 1364
(ii) (√3 + 1)5 = 5C0 (√3)5 + 5C1 (√3)3 + 5C3 (√3)2 + 5C4 √3 + 5C5 …(1)
and (√3 – 1)5 = 5C0 (√3)5 – 5C1 (√3)4 + 5C2 (√3)3 – 5C3 (√3)2 + 5C4 √3 – 5C5 …(2)
eqn. (1) – eqn. (2) gives ;
(√3 + 1)5 – (√3 – 1)5 = 2 × 5C1 (√3)4 + 2 × 5C3 (√3)2 + 2 × 5C5
= 2 × 5 × 9 + 2 × \(\frac{5 \times 4}{2}\) × 3 + 2 × 1 = 90 + 60 + 2 = 152 …(3)
Now from (3); (√3 + 1)5 – (√3 – 1)5 = 152
⇒ (√3 + 1)5 = 152 + (√3 – 1)5 > 152
[∵ (√3 – 1)5 > 0]
Also (√3 + 1)5 = 152 + (√3 – 1)5 < 153
[∵ (√3 – 1)5 < 1]
Hence (√3 + 1)5 lies between 152 and 153.
Question 6.
If the first three terms in the expansion of (1 + ax)n in ascending powers of x are 1 + 12x + 64x2, find n and a.
Solution:
By binomial Theorem, we have
(1 + ax)n =nC0 1n (ax)0 + nC1 (ax) + nC2 1n – 2(ax)2 + ….
= 1 + nax + \(\frac{n(n-1)}{2}\) a2x2 + ….
Also given first three terms of (1 + ax)n are 1 + 12x + 64x2
∴ na = 12 …(1)
and \(\frac{n(n-1)}{2}\)a2 = 64 …(2)
On squaring eqn. (1); we have
n2 a2 = 144 …(3)
On dividing eqn. (2) by eqn. (3); we have
\(\frac{n(n-1)}{2 n^2}\) = \(\frac{64}{144}\) ⇒ \(\frac{n-1}{2 n}\) = \(\frac{4}{9}\) ⇒ 9n – 9 = 8n ⇒ 9n – 8n = 9 ⇒ n = 9
∴ from (1); 9a = 12 ⇒ a = \(\frac{4}{9}\)
Question 7.
Find the first three terms in the expansion of [2 + x(3 + 4x)]5 in ascending powers of x.
Solution:
[2 + x(3 + 4x)]5 = [2 + 3x + 4x2]5 =(y + 4x2)5; where y = 2 + 3x
= 5C0 y5 (4x2)0 + 5C1 y4 (4x2)2 + 5C3 y2 (4x2)3 + 5C4 y(4x2)4 + 5C5 (4x2)5
= 5C0 (2 + 3x)5 + 5C1 (2 + 3x)4 (4x)2 + ….
= [5C0 25 + 5C1 24 × 3x + 5C2 23 × (3x)2 + 5C3 22 (3x)3 + 5C4 2 (3x)4 + 5C5 (3x)5] + 20x2 {4C0 24 + ….}
= [32 + 240x + 720x2 + …] + 20x2 {4C0 24 + …} + …..
= 32 + 240x + 1040x2 + ….
Question 8.
Expand (1 + 2x + 3x2)n in a series of ascending powers of x up to and including the term in x2.
Solution:
Question 9.
Write down the expansion by the binomial theorem of \(\left(3 x-\frac{y}{2}\right)^4\). By giving x and y suitable values, deduce the value of (29.5)4 correct to four significant figures.
Solution:
Putting x = 10 and y = 1 in eqn. (1); we get
(30 – 0.5)4 = (29.5)4 = 81 × 104 – 54 × (10)3 × 1 + \(\frac { 27 }{ 2 }\) × 102 × 12 – \(\frac { 3 }{ 2 }\) × 10 × 13 + \(\frac { 1 }{ 16 }\)
= 810000 – 54000 + 1350 – 15 + \(\frac { 1 }{ 6 }\) = 757300 [correct to 4 significant figures.]
Question 10.
Using binomial theorem, evaluate : (999)3.
Solution:
(999)3 = (1000 – 1)3
= 3C0 (1000)3 (-1)0 + 3C1(1000)2 (-1)1 + 3C2(1000) (-1)2 + 3C3 (1000)0 (-1)3
=1000000000 – 3000000 + 3000 – 1 = 997002999
Question 11.
Write down in terms of x and n, the term containing x3 in the expansion of \(\left(1-\frac{x}{n}\right)^n\) by the binomial theorem. If this term equals \(\frac { 7 }{ 8 }\) when x = -2, and n is a positive integer, calculate the value of n.
Solution:
By binomial theorem, we have
Question 12.
(i) Obtain the binomial expansion of (2 – √3)6 in the form a + b√3, where a and b are integers. State the corresponding result for the expansion (2 + √3)6 and
(ii) show that (2 – √3)6 is the reciprocal of (2 +√3)6.
Solution:
(i) By binomial Theorem, we have
(2 – √3)6 = 6C0 26 (- √3)0 + 6C1 25 (- √3) + 6C2 24 (-√3)2 + 6C3 23 (-√3)3 + 6C4 22 (-√3)4 + 6C5 21 (-√3)5 + 6C6 20 (-√3)6
= 1 × 64 + 6 × 32 (- √3) + \(\frac{6 \times 5}{2}\) × 16 × 3 + \(\frac{6 \times 5 \times 4}{6}\) × 8 × (-3√3) + \(\frac{6 \times 5}{2}\) × 4 × 9 + 6 × 2 × (-9√3) + 1 × 1 × 27
= 64 = 192√3 + 720 – 480√3 + 540 – 108√3 + 27
∴ (2 – √3)6 = 1351 – 780√3
Similarly (2 + √3)6 = 64 + 192√3 + 720 + 480√3 + 540 + 108√3 + 27 = 1351 + 780√3
(ii) Thus, (2 – √3)6 (2 + √3)6 = (1351 – 780√3)(1351 + 780√3) = (1351)2 – (780)2 × 3 = 1825201 – 1825, 200 = 1
⇒ (2 – √3)6 = \(\frac{1}{(2+\sqrt{3})^6}\)
Thus, (2 – √3)6 is the reciprocal of (2 + √3)6.
Question 13.
Find the coefficient of x5 in the expansion of (1 + 2x)6 (1 – x)7.
Solution:
Now using binomial theorem, we have
Question 14.
If the coefficients of second, third and fourth terms in the expansion of (1 + x)2n are in A.P., show that 2n2 – 9n + 7 = 0.
Solution:
By binomial theorem, we have
(1 + x)2n =2nC0 12n x0 + 2nC11 12n – 1 x + 2nC2 12n – 2 x2 + 2nC3 12n – 3 x3 + …..
∴ Coeff. of 2nd term = 2nC1
Coeff. of 3rd term = 2nC2
and Coeff. of 4 th term = 2nC3
Since it is given that coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)2n are in A.P
Question 15.
Let n be a positive integer. If the coefficients of 2nd, 3rd, 4th terms in the expansion of (1 + x)n are in A.P., then find the value of n.
Solution:
By binomial theorem, we have
(1 + x)n = nC0 1nx0 + nC1 1n – 1 x + nC2 1n – 2 x2 + nC3 1n – 3 x3 + …. + nCn 10 xn
∴ Coefficient of 2nd term = nC1
Coeff. of 3rd term = nC2
and Coeff. of 4th term = nC3
Since it is given that coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)n are in A.P.
Question 16.
In the binomial expansion of \((\sqrt[3]{3}+\sqrt{2})^5\) find the term which does not contain Irrational expression.
Solution:
Using binomial theorem, we have
