NCERT Exemplar Class 11 Maths Solutions Chapter 8 Binomial Theorem

Question 2

if the term free from x is the expansion of $\left ( \sqrt{x}-\frac{k}{x^{2}} \right )^{10}$ is 405, then find the value of k.

Answer:

$\left ( \sqrt{x}-\frac{k}{x^{2}} \right )^{10}$................... (Given)
$T_{r+1}=^{10}\textrm{C}_{r}\left ( \sqrt{x} \right )^{10-r}\left ( -\frac{k}{x^{2}} \right )^{r}$ ……. (from standard formula of Tr+1)

$\Rightarrow$$T_{r+1}=^{10}\textrm{C}_{r} x ^{\frac{10-5r}{2}}\left ( -k \right )^{r}$ …….. (i)
Now, for x,
(10 – 5r)/2 = 0
$\Rightarrow$ r = 2
Substituting the value of r in eq (i),
$T_{2+1}=^{10}\textrm{C}_{2} \left (-k \right )^{2}$= 405
$\Rightarrow$$\frac{10\times 9\times 8!}{2!\times 8!}\left ( k \right )^{2}=405$
$\Rightarrow$$45k^{2}=405$
Thus k² = 9
Thus, k = ±3

Question 3

Find the coefficient of x in the expansion of $\left ( 1-3x+7x^{2} \right )\left ( 1-x \right )^{16}$

Answer:

(1-3x+7x²) (1-x)¹⁶ ….. (given)
$\Rightarrow$ (1-3x+7x²) (¹⁰⁶C₀- ¹⁶C₁x¹ + ¹⁶C₂x² + …… + ¹⁶C₁₆x¹⁶)
$\Rightarrow$ (1-3x +7x²) (1-16x+120x²+…..)
Thus, coefficient of x = -19.

Question 4

Find the term independent of x in the expansion of $\left ( 3x-\frac{2}{x^{2}} \right )^{15}$

Answer:

Given: (3x – 2/x²)15
Thus, $T_{r+1} = ^{15}C_r3^{15-r}x^{15-3r}(-2)^r$
Now, for independent term x,
15 – 3r = 0
Thus, r = 5
Thus, the term x is
T₅₊₁ = ¹⁵C₅3^15-5(-2)⁵
$\Rightarrow$ $-\frac{15 × 14 × 13 × 12 × 11 × 10!}{5×4×3×2×1×10!}$ .3¹⁰.2⁵
$\Rightarrow$ -3003 × 3¹⁰ × 2⁵

Question 5

Find the middle term (terms) in the expansion of
i) $\left ( \frac{x}{a}-\frac{a}{x} \right )^{10}$ ii) $\left ( 3x-\frac{x^{3}}{6} \right )^{9}$

Answer:

(i) Given: (x/a – a/x)¹⁰
Where, n = 10(even).
The middle term = (10/2 + 1)th term, viz., the sixth term
Thus, T₆ = T₅₊₁¹⁰C₅ (x/a)^10-5 (-a/x)⁵
$\Rightarrow$ - ¹⁰C₅ (x/a)⁵(a/x)⁵
$\Rightarrow$ - 10 x 9 x 8 x 7 x 6 x 5!/ 5! x 4 x 3 x 2 x 1 (x/a)⁵(x/a)^-5
$\Rightarrow$ -252
(ii) Given: (3x – x³/6)⁹
Where, n = 9 (odd)
Now, here are two middle terms-
(9+1/2)^th ,viz., the tenth term
& (9+1/2 + 1)^th, viz., the sixth term
Thus, T₅ = T₄₊₁
= ⁹C₄ (3x)^9-4 (-x³/6)⁴
= 9x8x7x6x5!/4x3x2x1x5! . 3⁵x⁵x¹²6^-4
= 189/8 x¹⁷
Now,
T₆ = T₅₊₁
= ⁹C₅(3x)^9-5 (-x³/6)⁵
= - 9 x 8 x 7 x 6 x 5!/ 5! x 4 x 3 x 2 x 1 . 3⁴.x⁴.x¹⁵.6^-4
= - 21/16 x¹⁹

Question 6

Find the coefficient of x¹⁵ in the expansion of (x – x²)¹⁰

Answer:

Given: (x – x²)¹⁰
Thus, T_r+1 = ¹⁰C_r x^10-r(-x²)^r
= (-1)^{r 10}C_rx^10-rx^2r
= (-1)^{r 10}C_rx^10+r
Now, for the coefficient of x¹⁵
We have,
10 + r = 15
Thus, r = 5
Thus, T₅₊₁ = (-1)⁵.¹⁰C₅x¹⁵
Therefore,
Coefficient of x¹⁵ = -10x9x8x7x6x5!/5x4x3x2x1x5!
= -252

Question 7

Find the coefficient of x^1/17 in the expansion of (x⁴ – 1/x³)¹⁵

Answer:

Given: (x⁴ – 1/x³)¹⁵
Thus,
T_r+1= ¹⁵C_r(x⁴)^15-r(-1/x³)^r
$\Rightarrow$ ¹⁵C_r x^60-4r(-1)^r x^-3r
$\Rightarrow$ ¹⁵C_r x^60-7r(-1)^r
Now, for coefficient x^-17,
60 – 7r = -17
$\Rightarrow$7r = 77
$\Rightarrow$r = 11
Thus, T₁₁₊₁ = ¹⁵C₁₁ x^60-77(-1)¹¹
Therefore,
Coefficient of x^-17 = -15 x 14 x 13 x 12 x 11!/11! x 4 x 3 x 2 x 1
= -1365

Question 8

Find the sixth term of the expansion (y^1/2 + x^1/3)ⁿ, if the binomial coefficient of the third term from the end is 45.
Answer:

Given: (y^1/2 + x^1/3)ⁿ
In the end, the binomial coefficient of 3^rd term = 45
Thus, ⁿC_n-2 = 45 , ⁿC₂= 45
i.e., n(n-1)(n-2)!/2!(n-2)! = 45
Now, n(n-1) = 90,
$\Rightarrow$n² – n – 90 = 0,
$\Rightarrow$(n – 10)(n + 9) = 0
$\Rightarrow$n = 10 …. (since n is not equal to -9)
Now, the 6^th term,
=¹⁰C₅ (y^1/2)^10-5(x^1/3)⁵
= 252y^5/2.x^5/3

Question 9

Find the value of r, if the coefficients of (2r + 4)^th and (r – 2)^th terms in the expansion of (1 + x)¹⁸ are equal.

Answer:

Given: (1+x)¹⁸
Now, T_(2r+3)+1is the (2r + 4)^th term
Thus, T_(2r+3)+1 = ¹⁸C_2r+3 (x)^2r+3
Now, T_(r-3)+1 is the (r-2)^th term
Thus, T_(r-3)+1 = ¹⁸C_r-3 x^r-3
Now, it is given that the terms are equal in expansion,
Thus,
¹⁸C_2r+3= ¹⁸C_r-3
i.e., 2r + 3 + r – 3 = 18 …… (ⁿC_x = ⁿC_y gives x + y = n)
Thus, 3r = 18
Therefore, r = 6

Question 10

If the coefficients of the second, third and fourth terms in the expansion of (1+x)²ⁿ are in A.P., then show that 2n² – 9n + 7 = 0.
Answer:

Given: (1+x)²ⁿ
Now, ²ⁿC₁, ²ⁿC₂, ²ⁿC₃ are the coefficients of the first, second, and third terms, respectively.
It is also given that the coefficients are in A.P.
Thus, 2. ²ⁿC₂=²ⁿC₁ +²ⁿC₃
Thus, 2[2n(2n-1)(2n-2)!/2x1x(2n-2)!] = 2n + 1n(2n-1)(2n-2)(2n-3)!/3!(2n-3)!
$\Rightarrow$n(2n-1) = n = n(2n-1)(n-1)/3
$\Rightarrow$3(2n-2) = 3+ (2n² – 3n + 1)
$\Rightarrow$6n – 3 = 2n² – 3n + 4
Thus, 2n² – 9n + 7 = 0

Question 11

Find the coefficient of x⁴ in the expansion of (1 + x + x² + x³)^11.

Answer:

Given: (1+x+x²+x³)¹¹
That is equal to [(1+x) + x²⁽1 + x)]¹¹
$\Rightarrow$ [(1+x) (1+x²)]¹¹
$\Rightarrow$ (1+x)¹¹.(1+x²)¹¹
$\Rightarrow$ (¹¹C₀ + ¹¹C₁x + ¹¹C₂x² + ¹¹C₃x³ +¹¹C₄x⁴ + ….) (_¹¹C₀ + ¹¹C₁x² + ¹¹C₂x4 + ….)
Thus, the coefficient of x⁴ =¹¹C₀ x ¹¹C4 + ¹¹C1 x ¹¹C2+¹¹C2x ¹¹C₀
= 330 + 605 + 55
= 990

Question 12

If p is a real number and the middle term in the expansion (p/2 + 2)⁸ is 1120, then find the value of p.

Answer:

Given: (p/2 + 2)⁸
Now, here the index is n = 8,
Since there is only one middle term, viz., (8/2 + 1), i.e., the fifth term
T₅ = T₄₊₁
= ⁸C₄(p/2)^8-4. 2⁴
Now,
1120 = ⁸C₄p⁴
$\Rightarrow$1120 = 8x7x6x5x4!/4!x4x3x2x1 . p⁴
$\Rightarrow$1120 = 7x2x5xp⁴
Thus, p⁴= 1120/70
$\Rightarrow$p⁴= 16
$\Rightarrow$p² = 4
Thus, p = ±2

Question 13

Show that the middle term in the expansion of$\left ( x-\frac{1}{x} \right )^{2n}$is$\frac{1\times 3 \times 5\times...........\times\left ( 2n-1 \right ) }{n!} \times\left ( -2 \right )^{n}$
Answer:

Given: (x – 1/x)²ⁿ
Now, here the index = 2n(even)
There’s only one middle term, i.e., (2n/2 + 1)^thterm, viz., (n + 1)^thterm
T_n+1 = ²ⁿC_n(x)^2n-n(-1/x)ⁿ
$\Rightarrow$ ²ⁿC_n(-1)ⁿ
$\Rightarrow$ (-1)ⁿ(2n!)/n!.n!
$\Rightarrow$ (-1)ⁿ 1.2.3.4.5. …. (2n-1).(2n)/n!.n!
$\Rightarrow$ (-1)ⁿ [1.3.5…. (2n-1)].2ⁿ[1.2.3…..n]/(1.2.3…n).n!
$\Rightarrow$ (-2)ⁿ[1.3.5…(2n-1)].2ⁿ/n!

Question 14

Find n in the binomial $\left (\sqrt[3]{2} +\frac{1}{\sqrt[3]{3}} \right )^{n}$ if the ratio of 7th term from the beginning to the 7th term from the end is $\frac{1}{6}$

Answer:

Given $\left ( \sqrt[3]{2}+\frac{1}{\sqrt[3]{3}} \right )^{n}$
Now, from the beginning,
Seventh term is $T_{7}=T_{6+1}$
$=^{n}C_{6} \left (\sqrt[3]{2} \right )^{n-6}\left ( \frac{1}{\sqrt[3]{3}} \right )^{6}$...................(i)
& from the end,
The seventh term is the same ,i.e., $\left (\frac{1}{\sqrt[3]{3}}+\sqrt[2]{3} \right )^{n}$
$T_{7}=^{n}C_{6}\left ( \frac{1}{\sqrt[3]{3}} \right )^{n-6}\left ( \sqrt[3]{2} \right )^{6}$.............(ii)
Now, it is given that
$\frac{^{n}C_{6}\left (\sqrt[3]{2} \right )^{n-6}\left ( \frac{1}{\sqrt[3]{3}} \right )^{6}}{^{n}C_{6}\left ( \frac{1}{\sqrt[3]{3}} \right )^{n-6}\left ( \sqrt[3]{2} \right )^{6}} = \frac{1}{6}$
Thus
$\frac{\left ( \sqrt[3]{2} \right )^{n-12}}{\left (\frac{1}{\sqrt[3]{3}} \right )^{n-12} }=\frac{1}{6}$
Thus $(\sqrt[3]{2} \sqrt[3]{3})^{n-12}=6^{-1}$

Thus $6^\frac{n-12}{3}=6^{-1}$
$\frac{n-12}{3}=-1$
Thus n=9

Question 15

In the expansion of (x + a)n, if the sum of odd terms is denoted by O and the sum of even terms by E.
Then prove that
(i) O²– E²= (x²– a²)n (ii) 4OE = (x + a)²ⁿ- (x - a)²ⁿ

Answer:

(i) (x+a)ⁿ= ⁿC₀xⁿ + ⁿC₁x^n-1 a¹ + ⁿC₂x^n-2 a² + ⁿC₃x^n-3 a³ + ….. + ⁿC_n aⁿ
Now, the sum of odd terms,
O = ⁿC₀xⁿ+ ⁿC₂x^n-2 a² + …….
& sum of even terms,
E = ⁿC₁x^n-1 a¹ + ⁿC₃x^n-3 a³ + …..
Now, (x+a)ⁿ = O+E & (x-a)ⁿ = O – E
Thus, (O + E)(O – E) = (x + a)ⁿ(x – a)ⁿ
Thus, O² – E² = (x² – a²)ⁿ
(ii)Here,
4OE = (O + E)² – (O – E)²
$\Rightarrow$ [(x+a)ⁿ]² – [(x-a)ⁿ]²
$\Rightarrow$ (x+a)²ⁿ – (x-a)²ⁿ

Question 16

If xⁿ occurs expantion of $\left (x^{n}+\frac{1}{x} \right )^{2n}$ then prove that its coefficient is $\frac{\left ( 2n \right )!}{\left ( \frac{4n-p}{3} \right )!\left ( \frac{2n+p}{3} \right )!}$

Answer:

Given: (x² + 1/x)²ⁿ
Thus, T_r+1=²ⁿC_r(x²)^2n-r(1/x)^r
= ²ⁿC_rx^4n-3r
Now, if xn will occur in the expansion, then,
Let us consider 4n – 3r = p
Thus,
$\Rightarrow$3r = 4n – p
$\Rightarrow$r = 4n – p/3
Thus, the coefficient of xp = ²ⁿC_r
$\Rightarrow$ (2n)!/r!(2n-r)!
$\Rightarrow$ $\frac{\left ( 2n \right )!}{\left ( \frac{4n-p}{3} \right )!\left ( \frac{2n+p}{3} \right )!}$

Question 17

Find the term independent of x in the expansion of(1+x+2x³)(3x²/2 – 1/3x)⁹

Answer:

Given: (1+x+2x³)( 3x²/2 – 1/3x)⁹
Now, let us consider (3x²/2 – 1/3x)⁹
T_r+1= ⁹C_r(3/2 x²)^9-r (-1/3x)^r
= ⁹C_r (3/2)^9-r (-1/3)^rx^18-3r
Thus, in the expansion of (1+x+2x³)(3/2x² – 1/3x)⁹ the general term is:
⁹C_r(3/2)^9-r (-1/3)^r x^18-3r+ ⁹C_r (3/2)^9-r(-1/3)^r x^19-3r + 2.⁹C_r (3/2)^9-r(-1/3)^r x^21-3r
Now, for the independent term x,
Let us put 18 – 3r = 0,
19 – 3r = 0
& 21 – 3r = 0,
We get,
R = 6 & r = 7
Now, since the second term is not independent of x, it will be:
⁹C₆ (3/2)^9-6 (-1/3)⁶ + 2.⁹C_r (3/2)^9-7 (-1/3)⁷
= 9x8x7x6!/6!x3x2 . 1/2³.3³ – 2. 9x8x7!/7!x2x1. 3²/2².1/3⁷
= 84/8.1/3³ – 36/4.2/3⁵
= 17/54

Question 18

The total number of terms in the expansion of (x + a)¹⁰⁰+ (x – a)¹⁰⁰after simplification is
(A) 50 (B) 202 (C) 51 (D) none of these

Answer:

(c) 51.
Now, (x+a)¹⁰⁰+ (x-a)¹⁰⁰ = (¹⁰⁰C₀x¹⁰⁰ + ¹⁰⁰C₂x⁹⁹a + ¹⁰⁰C₂x⁹⁸a² + ….) + (¹⁰⁰C₀x¹⁰⁰- ¹⁰⁰C₂x⁹⁹a + ¹⁰⁰C₂x⁹⁸a²)
= 2(¹⁰⁰C₀x¹⁰⁰ + ¹⁰⁰C₂x⁹⁸a²+…..+¹⁰⁰C₁₀₀a¹⁰⁰)
Therefore, there are 51 terms.

Question 19

Given the integers r > 1, n > 2, and coefficients of (3r)^th and (r + 2)^nd terms in the binomial expansion of (1 + x)²ⁿ are equal, then
(A) n = 2r (B) n = 3r (C) n = 2r + 1 (D) none of these

Answer:

(a) n =2r
Given: (1+x)²ⁿ
Thus, T_3r = T_(3r-1)+1
= ²ⁿC_3r-1x^3r-1
& T_r+2= T_(r+1)+1
= ²ⁿC_r+1x^r+1
Now, ²ⁿC_3r-1x^3r-1 = ²ⁿC_r+1x^r+1 ……… (given)
Thus, 3r – 1 + r + 1 = 2n
Thus, n = 2r

Question 20

The two successive terms in the expansion of (1 + x)²⁴ whose coefficients are in the ratio 1: 4 are
(A) 3^rd and 4^th (B) 4^th and 5^th (C) 5^th and 6^th (D) 6^th and 7^th

Answer:

(c) 5^th & 6^th
Let us consider (r+1)^th& (r+2)^th as the 2 successive terms in the expansion of (1+x)²⁴
Now, T_r+1 = ²⁴C_rx^r& T_r+2 = ²⁴C_r+1x^r+1
Now, ²⁴C_r/²⁴C_r+1 = ¼ ……. (given)
Thus,
$\frac{\frac{\left ( 24 \right )!}{r!\left ( 24-r \right )!}}{\frac{\left ( 24 \right )!}{\left ( r+1 \right )\left ( 24-r-1 \right )!}}=\frac{1}{4}$
Thus, (r+1)r! (23 – r)! / r!(24 – r)(23 – r)! = ¼
$\Rightarrow$r + 1/24 – r = ¼
$\Rightarrow$4r + 4 = 24 – r
$\Rightarrow$r = 4
Thus, T₄₊₁ = T₅
& T₄₊₂ = T₆
Thus, opt (c).

Question 21

The coefficients of xⁿ in the expansion of (1 + x)²ⁿ and (1 + x)^{2n ~1} are in the ratio
(a) 1 : 2
(b) 1 : 3
(c) 3 : 1
(d) 2:1

Answer:

The answer is the option (d) 2:1
²ⁿC_n is the coefficient of xⁿ in the expansion of (1+x)²ⁿ
&^2n-1C_n is the coefficient of xⁿ in the expansion of (1+x)^2n-1
$\frac{^{2n}C_n}{^{2n-1}C_n}=\frac{\frac{\left ( 2n \right )!}{\left ( n \right )!n!}}{\frac{\left ( 2n-1 \right )!}{n!\left ( n-1 \right )!}}$
$=\frac {(2n)!n!(n-1)!}{(2n-1)!n!n!}$
$= \frac{2n(2n-1)!n!(n-1)!}{n!n(n-1)!(2n-1)!}$
$=\frac 21$
$= 2:1$

Question 22

If the coefficients of the 2^nd, 3^rd and 4^th terms in the expansion of (1 + x)ⁿ are in A.P., then the value of n is
(a) 2
(b) 7
(c) 11
(d) 14

Answer:

The answer is option (b) 7
(1+x)ⁿ = ⁿC₀+ ⁿC₁x + ⁿC₂x² +ⁿC₃x³ + ….. + ⁿC_nxⁿ
Now, ⁿC₁, ⁿC₂, ⁿC₃ are the coefficients of the second, third, and fourth terms, respectively.
2 ⁿC₂ =ⁿC₁ + ⁿC₃ …… since they are in A.P.
$\Rightarrow 2[\frac{n!}{(n-2)!2!}] = \frac{n + n!}{3!(n-3)!}$
$\Rightarrow2[\frac{n(n-1)}{2!}] = \frac{n + n(n-1)(n-2)}{3}!$
$\Rightarrow (n-1) = \frac{1 + (n-1)(n-2)}{6}$
$\Rightarrow$ 6n – 6 = 6 + n² – 3n + 2
$\Rightarrow$ n² – 9n + 14 = 0
$\Rightarrow$ (n-7)(n-2) = 0
Thus, n = 2 or n = 7
Now, n = 2 is not possible,
Thus, n = 7

Question 23

If A and B are coefficients of xⁿ in the expansions of (1 + x)²ⁿ and (1 + x)^2n–1 respectively, then A/B equals to
(a) 1
(b) 2
(c) $\frac{1}{2}$
(d)$\frac{1}{n}$

Answer:

The answer is the option (b) 2
²ⁿC_{n i}s the coefficient of xn in the expansion of (1+x)²ⁿ
Thus, A = ²ⁿC_n
&^2n-1C_n is the coefficient of xⁿ in the expansion of (1+x)^2n-1
Thus, B =^2n-1C_n
Thus, A/B = ²ⁿC_n/^2n-1C_n
= 2/1
=2:1

Question 24

If the middle term of $\left ( \frac{1}{x} + x \sin x \right )^{10}$ is equal to $7\tfrac{7}{8}$ then the value of x is
(a) $2n\pi + \frac{\pi}{6}$ (b) $n\pi + \frac{\pi}{6}$ (c) $n\pi +\left ( -1 \right )^{n} \frac{\pi}{6}$ (d) $n\pi +\left ( -1 \right )^{n} \frac{\pi}{3}$

Answer:

The answer is the option (c) nπ + (-1)ⁿπ/6
Now, n = 10 (even)
Thus, there is only one middle term i.e., the 6^th term
Thus, T₆ = T₅₊₁
= ¹⁰C₅(1/x)^10-5 (x sin x)⁵
$\Rightarrow$63/8 = ¹⁰C₅sin⁵x …… (given)
$\Rightarrow$63/8 = 252 x sin⁵x
Thus, sin⁵x = 1/32
$\Rightarrow$Sin x = ½
$\Rightarrow$Sin x = π/6
Thus, x = nπ + (-1)ⁿπ/6

Question 25

The largest coefficient in the expansion of (1 + x)³⁰ is _________________ .
Answer:

³⁰C_30/2 = ³⁰C₁₅ is the largest coefficient in the expansion of (1+x)³⁰

Question 27

In the expansion of (x² – 1/x²)¹⁶, the value of constant term is_________________.

Answer:

Here, T_r+1= ¹⁶C_r(x²)^16-r(-1/x²)^r
= ¹⁶C_rx^32-4r(-1)^r
Now, for the constant term,
32 – 4r = 0
Thus, r = 8
Thus, T₈₊₁ = ¹⁶C₈

Question 28

If the seventh terms from the beginning and the end in the expansion of $\left ( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}}\right )^{n}$ are equal, then n equals _________________

Answer:

Given: $\left ( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \right )^{n}$
Now, referring to que 14.
$\frac{^{n}C_{6} \left ( \sqrt[3]{2} \right )^{n-6}\left ( \frac{1}{\sqrt[3]{3}} \right )^{6}}{^{n}C_{6}\left ( \frac{1}{\sqrt[3]{3}} \right )^{n-6}\left ( \sqrt[3]{2} \right )^{6}}=1$
Thus
$\frac{\left ( \sqrt[3]{2} \right )^{n-12}}{\left (\frac{1}{\sqrt[3]{3} }\right )^{n-12}}=1$
Thus
6^(n-12)/3 = 6⁰
(n-12)/3 = 0
Thus, n = 12

Question 29

The coefficient of $a^{-6}b^{4}$ in the expansion of $\left ( \frac{1}{a}-\frac{2b}{3} \right )^{10}$ is ______________

Answer:

Given: (1/a – 2b/3)¹⁰
Thus, T_r+1= ¹⁰C_r(1/a)^10-r (-2b/3)^r
Now, for the coefficient of a^-6b⁴,
10 – r = 6
Thus, r = 4
Therefore the coefficient of a^-6b⁴ = ¹⁰C₄(-2/3)⁴
= 10.9.8.7.6!/6!.4.3.2.1 . 2⁴/3⁴
= 1120/27

Question 30

Middle term in the expansion of (a³ + ba)²⁸is _________

Answer:

Given: (a³ + ba)²⁸
Here, n = 28 (even)
Thus, there is only one middle term (28/2 + 1)th or 15^th term.
Thus, the middle term, T₁₅ = T₁₄₊₁
$\Rightarrow$ ²⁸C₁₄(a³)^28-14(ba)¹⁴
$\Rightarrow$ ²⁸C₁₄a⁴²b¹⁴a¹⁴
$\Rightarrow$ ²⁸C₁₄a⁵⁶b¹⁴

Question 31

The ratio of the coefficients of x^p and x^q in the expansion of (1 +x)^p+q is _______

Answer:

Given: (1+x)^p+q
Thus, coefficient of x^p = ^p+qC_p
& Coefficient of x^q = ^p+qC_q
Therefore,
^p+qC_p/ ^p+qC_q = ^p+qC_p /^p+qC_p
= 1:1

Question 32

The position of the term independent of x in the expansion of $\left ( \sqrt{\frac{x}{3}} + \frac{3}{2x^{2}}\right )^{10}$ is _____________.

Answer:

Given: $\left ( \sqrt{\frac{x}{3}}+\frac{3}{2x^{2}} \right )^{10}$
Thus, $T_{r+1}=^{10}C_{r}\left ( \sqrt{\frac{x}{3}} \right )^{10-r}\left ( \frac{3}{2x^{2}} \right )^{r}$
Now, for constant term,
10 – 5r = 0
Thus, r = 2
Therefore, the third term is independent of x.

Question 33

If 25¹⁵ is divided by 13, the reminder is _________ .

Answer:

Now, 25¹⁵= (26 – 1)¹⁵
= ¹⁵C₀26¹⁵– ¹⁵C₁26¹⁴ + …. + ¹⁵C₁₄26 – ¹⁵C₁₅
= (¹⁵C₀26¹⁵ – ¹⁵C₁26¹⁴ + …. + ¹⁵C₁₄26 – 13) + 12
Therefore, it is clear that, when we divide 25¹⁵by 13, we get 12 as the remainder.

Question 34

The sum of the series $\sum_{10}^{r=0} 20 \mathrm{C}_r$ is $2^{19}+\frac{{ }^{20} C_{10}}{2}$.

Answer:

The given statement is False.
Now, from what’s given, we can say that,
$\Rightarrow$ $\sum_{10}^{r=0} 20 \mathrm{C}_r$ = ²⁰C₀ + ²⁰C₁ + ²⁰C₂ +²⁰C₃ + …… + ²⁰C₁₀
$\Rightarrow$ ²⁰C₀+ ²⁰C₁ + ….. + ²⁰C₁₀ +²⁰C₁₁ + …… + ²⁰C₂₀ – (²⁰C₁₁ + ….. + ²⁰C₂₀)
$\Rightarrow$ 2²⁰– (²⁰C₁₁ + …… + ²⁰C₂₀)

Question 35

The expression 7⁹ + 9⁷ is divisible by 64.

Answer:

The given statement is True.
Given: 7⁹ + 9⁷ is divisible by 64.
Now,
7⁹ + 9⁷ = (1+8)⁷ – (1-8)⁹
$\Rightarrow$ [⁷C₀+⁷C₁.8 + ⁷C₂.(8)² + ⁷C₃.(8)³ + ….. + ⁷C₇.(8)⁷] – [⁹C₀ + ⁹C₁.8 + ⁹C₂.(8)² - ⁹C₃.(8)³ + ….. ⁹C₉.(8)⁰]
$\Rightarrow$ (7x8 + 9x8) + (21 x 8² – 36 x 8²) + …..
$\Rightarrow$ 128 + 64(21 – 36)
$\Rightarrow$ 64[2 + (21 – 36) + …], viz. divisible by 64.

Question 36

The number of terms in the expansion of [(2x + y³)⁴]⁷ is 8.

Answer:

The given statement is False.
Given: [(2x + y³)⁴]⁷ = (2x + y³)²⁸
Thus, no. of terms = 28 + 1
= 29

Question 37

The sum of coefficients of the two middle terms in the expansion of (1+x)^2n-1 is equal to ^{2n – 1}C_n.

Answer:

The given statement is False.
Given: (1 + x)^2n-1
Now, no. of terms = 2n – 1 + 1
= 2n (even)
Thus, the middle term = 2n/2^thterms & (2n/2 + 1)^th terms
= n^thterms & (n+1)^th terms
Now, the coefficient of-
n^th term =^{2n – 1}C_n-1
& (n + 1)^th term = ^2n-1C_n
Now, sum of the coefficients = ^{2n – 1}C_n-1+ ^2n-1C_n
= ^2n-1C_n-1 + ^2n-1C_n
=²ⁿC_n

Question 38

The last two digits of the number 3⁴⁰⁰ are 01.

Answer:

The given statement is true.
Given: 3⁴⁰⁰= (9)²⁰⁰
= (10 – 1)²⁰⁰
Thus, (10 – 1)²⁰⁰ = ²⁰⁰C₀(10)²⁰⁰ – ²⁰⁰C₁(10)¹⁹⁹+ ….. – ²⁰⁰C₁₉₉(10)¹ +²⁰⁰C₂₀₀(1)²⁰⁰
= 10²⁰⁰ – 200 x 10¹⁹⁹ + …. – 10 x 200 + 1
Therefore, the last two digits will be 01.

Question 39

If the expansion of (x-1/x²)²ⁿ contains a term independent of x, then n is a multiple of 2.

Answer:

The given statement is False.
Given: $\left ( x-\frac{1}{x^{2}} \right )^{2n}$
Now,
$\Rightarrow$ T_r+1 = ²ⁿC_r(x)^2n-r (-1/x²)^r
$\Rightarrow$ ²ⁿC_r(x)^2n-r-2r(-1)^r
$\Rightarrow$ ²ⁿC_r(x)^2n-3r (-1)^r(-1)^r
Now, for the independent term of x,
2n – 3r = 0
Thus, r = 2n/3, viz. not an integer
Thus, this expression is not possible to be true.

Question 40

Numbers of terms in the expansion of (a + b)ⁿ, where n ∈ N, is one less than the power n.

Answer:

The given statement is False.
In the given expression (a+b)ⁿ, the no. of terms is just 1 more than n, i.e., n + 1.