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Maths Binomial Theorem Maths Class 11th Binomial Theorem

If ²⁰C_r is the co-efficient of x^r in the expansion of (1 + x)²⁰, then the value of $\sum_{r = 0}^{20} r^{2}$ ²⁰C_r is equal to

Option 1 -

380 * 2¹⁸

Option 2 -

420 * 2¹⁸

Option 3 -

420 * 2¹⁹

Option 4 -

380 * 2¹⁹

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Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

Kindly consider the following figure

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Similar Questions for you

If the coefficient of x¹⁰ in the binomial expansion of ${(\frac{\sqrt[]{x}}{5^{\frac{1}{4}}} + \frac{\sqrt[]{5}}{x^{\frac{1}{3}}})}^{60}$ is $5^{k} . l,$ where $l, k \in N a n d l$ is co-prime to 5, then k is equal to………...

alok kumar singh

$T_{r + 1} =^{60} C_{r} {(x^{\frac{1}{2}})}^{60 - r} {(x^{- \frac{1}{3}})}^{r} {(5^{\frac{- 1}{4}})}^{60 - r} {(5^{\frac{1}{2}})}^{r}$

for

$x^{10} \frac{60 - r}{2} - \frac{r}{3} = 10$

$\Rightarrow 180 - 3 r - 2 r = 60$

->r = 24

k = 3 + exponent of 5 in

= $3 + ([\frac{60}{5}] + [\frac{60}{5^{2}}] - [\frac{24}{5}] - [\frac{24}{5^{2}}] - [\frac{36}{5}] - [\frac{36}{5^{2}}])$

= 3 + (12 + 2 – 4 – 0 – 7 – 1)

= 3 + 2 = 5

15. Find the expansion of (3x²– 2ax + 3a²)³using binomial theorem.

Payal Gupta

15. ${[3 x^{2} - 2 a x + 3 a^{2}]}^{3}$

= ${[3 x^{2} + a (- 2 x + 3 a)]}^{3}$

We know that by binomial theorem,

${(a + b)}^{3}$ = $a^{3} + b^{3} + 3 a b (a + b)$

= $a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}$

Then,

${[3 x^{2} + a (- 2 x + 3 a)]}^{3}$

= (3x²)³ + ${[a (- 2 x + 3 a)]}^{3}$ + $[3 (3 x^{2})^{2} a (- 2 x + 3 a)]$ + $[3 (3 x^{2}) {a (- 2 x + 3 a)}^{2}]$

= 27x⁶ + $[a^{3} {(- 2 x + 3 a)}^{3}]$ + $[3 (9 x^{4}) (- 2 a x + 3 a^{2})]$ + $[3 (3 x^{2}) {a^{2} {(3 a - 2 x)}^{2}}]$

= 27x⁶ + $[a^{3} {- 8 x^{3} + 27 a^{3} + 3 (4 x^{2}) (3 a) + 3 (- 2 x) (9 a^{2})}]$ + $[- 54 a x^{5} + 81 a^{2} x^{4}]$ + [ $(9 a^{2} x^{2})$ $(9 a^{2} + 4 x^{2} - 12 a x)$ ]

= 27x⁶ + [ $- 8 a^{3} x^{3} + 27 a^{6} + 36 a^{4} x^{2} - 54 a^{5} x$ ] + [ $- 54 a x^{5} + 81 a^{2} x^{4}$ ] + [ $(81 a^{4} x^{2} + 36 a^{2} x^{4} - 108 a^{3} x^{3}$ ]

= 27x⁶ $- 8 a^{3} x^{3} + 27 a^{6} + 36 a^{4} x^{2} - 54 a^{5} x$ $- 54 a x^{5} + 81 a^{2} x^{4}$ + $81 a^{4} x^{2} + 36 a^{2} x^{4} - 108 a^{3} x^{3}$

= 27x⁶– 54ax⁵ $+ 117 a^{2} x^{4}$ $- 116 a^{3} x^{3}$ + $117 a^{4} x^{2}$ $- 54 a^{5} x$ $+ 27 a^{6}$

14. 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]

Payal Gupta

14. For (a – b) to be a factor of aⁿ – b ⁿwe need to show (aⁿ – bⁿ) = (a – b)k as k is a natural number.

We have, for positive n

aⁿ = ${(a - b + b)}^{n}$ = ${[(a - b) + b]}^{n}$

=>aⁿ = ⁿC₀(a – b)ⁿ + ⁿC₁(a – b)^{n -}¹b + ⁿC₂(a – b)^{n –}²b² + ………… +ⁿC_n_-1 $(a - b) b^{n - 1}$ + ⁿC_nbⁿ

=>aⁿ= ${(a - b)}^{n}$ + ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + …………….…+ ⁿC_n_-1 $(a - b) b^{n - 1}$ + $b^{n}$ [Since, ⁿC₀ = 1 and ⁿC_n = 1]

=> $a^{n} - b^{n}$ = ${(a - b)}^{n}$ +ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + ……………… + ⁿC_n_-1 $(a - b) b^{n - 1}$

=> $a^{n} - b^{n}$ = $(a - b)$ [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +………..…… + ⁿC_n_-1 $b^{n - 1}$ ]

=> $a^{n} - b^{n}$ &n

...more

14. For (a – b) to be a factor of aⁿ – b ⁿwe need to show (aⁿ – bⁿ) = (a – b)k as k is a natural number.

We have, for positive n

aⁿ = ${(a - b + b)}^{n}$ = ${[(a - b) + b]}^{n}$

=>aⁿ = ⁿC₀(a – b)ⁿ + ⁿC₁(a – b)^{n -}¹b + ⁿC₂(a – b)^{n –}²b² + ………… +ⁿC_n_-1 $(a - b) b^{n - 1}$ + ⁿC_nbⁿ

=>aⁿ= ${(a - b)}^{n}$ + ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + …………….…+ ⁿC_n_-1 $(a - b) b^{n - 1}$ + $b^{n}$ [Since, ⁿC₀ = 1 and ⁿC_n = 1]

=> $a^{n} - b^{n}$ = ${(a - b)}^{n}$ +ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + ……………… + ⁿC_n_-1 $(a - b) b^{n - 1}$

=> $a^{n} - b^{n}$ = $(a - b)$ [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +………..…… + ⁿC_n_-1 $b^{n - 1}$ ]

=> $a^{n} - b^{n}$ = $(a - b)$ k where k = [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +………..…… + ⁿC_n_-1 $b^{n - 1}$ ] is a natural number.

Therefore (a – b) is a factor of aⁿ – bⁿwhere n is positive integer.

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13. Find a if the coefficients of x²and x³in the expansion of (3 + ax)⁹are equal.

Payal Gupta

13. The general term of the expansion ${(3 + a x)}^{9}$ is

T_r₊₁ = ⁹C_r $3^{(9 - r)}$ ${(a x)}^{r}$

= ⁹C_r $3^{(9 - r)} a^{r} x^{r}$

At r = 2,

T₂₊₁ = ⁹C₂ $3^{(9 - 2)} a^{2} x^{2}$

= $\frac{9!}{2! (9 - 2)!}$ 3⁷a²x²

= $9′$ $8′$ $7!$ / $2′$ $1′$ $7!$ 3⁷a²x²

= 36 ×3⁷a²x²

At r = 3,

T₃₊₁ = ⁹C₃ $3^{(9 - 3)} a^{3} x^{3}$

= $\frac{9!}{3! (9 - 3)!}$ 3⁶a³x³

= 9'8'7'6!/3'2'1'6! 3⁶a³x³

= 84 ×3⁶a³x³

Given that,

Co-efficient of $x^{2}$ = co-efficient of $x^{3}$

=> 36 × $3^{7} a^{2}$ = 84 × $3^{6} a^{3}$

=> $\frac{a^{3}}{a^{2}}$ = 36' 3⁷^/84'3⁶

=> $a$ = $3′3/7$

= $\frac{9}{7}$

12. 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.

Payal Gupta

1.The general term of the expansion (a + b)ⁿ is given by

T_r₊₁ = ⁿC_raⁿ^–rb^r

So, T₁ = ⁿC₀aⁿ = aⁿ

T₂ = ⁿC₁aⁿ^-1b = $\frac{n!}{1! (n - 1)!}$ aⁿ^-1 b = $\frac{n \times (n - 1)!}{(n - 1)!}$ aⁿ^-1b = naⁿ^-1b

T₃ = ⁿC₂aⁿ^-2b² = $\frac{n!}{2! (n - 2) [!}$ aⁿ^-2b² = $\frac{n \times (n - 1) \times (n - 2)!}{2 \times 1 \times (n - 2)!}$ aⁿ^-2b² = $\frac{n (n - 1)}{2}$ aⁿ^-2b²

Given,

T₁ = 729

=>aⁿ = 729 ------------------ (1)

T₂ = 7290

=>naⁿ^–1b = 7290 ------------- (2)

T₃ = 30375

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b² = 30375 ------------------- (3)

Dividing equation (2) by (1) we get,

$\frac{n a^{n - 1} b}{a^{n}}$ = $\frac{7290}{729}$

=> $\frac{n b}{a}$ = 10

Similarly dividing equation (3) by (2) we get,

$\frac{n (n - 1)}{2}$ aⁿ^–2b² ÷ naⁿ^–1b = $\frac{30375}{7290}$

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b²× $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$

=> $n (n - 1) a^{n - 2} b^{2}$ × $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$ × 2

=> $\frac{(n - 1) b}{a}$ = $\frac{2 5′ \times 2}{6}$

=> $\frac{n b}{a}$ $- \frac{b}{a}$ =

...more

1.The general term of the expansion (a + b)ⁿ is given by

T_r₊₁ = ⁿC_raⁿ^–rb^r

So, T₁ = ⁿC₀aⁿ = aⁿ

T₂ = ⁿC₁aⁿ^-1b = $\frac{n!}{1! (n - 1)!}$ aⁿ^-1 b = $\frac{n \times (n - 1)!}{(n - 1)!}$ aⁿ^-1b = naⁿ^-1b

T₃ = ⁿC₂aⁿ^-2b² = $\frac{n!}{2! (n - 2) [!}$ aⁿ^-2b² = $\frac{n \times (n - 1) \times (n - 2)!}{2 \times 1 \times (n - 2)!}$ aⁿ^-2b² = $\frac{n (n - 1)}{2}$ aⁿ^-2b²

Given,

T₁ = 729

=>aⁿ = 729 ------------------ (1)

T₂ = 7290

=>naⁿ^–1b = 7290 ------------- (2)

T₃ = 30375

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b² = 30375 ------------------- (3)

Dividing equation (2) by (1) we get,

$\frac{n a^{n - 1} b}{a^{n}}$ = $\frac{7290}{729}$

=> $\frac{n b}{a}$ = 10

Similarly dividing equation (3) by (2) we get,

$\frac{n (n - 1)}{2}$ aⁿ^–2b² ÷ naⁿ^–1b = $\frac{30375}{7290}$

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b²× $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$

=> $n (n - 1) a^{n - 2} b^{2}$ × $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$ × 2

=> $\frac{(n - 1) b}{a}$ = $\frac{2 5′ \times 2}{6}$

=> $\frac{n b}{a}$ $- \frac{b}{a}$ = $\frac{25}{3}$

=> 10 – $\frac{b}{a}$ = $\frac{25}{3}$ [since, $\frac{n b}{a}$ = 10]

=> $\frac{b}{a}$ = 10 – $\frac{25}{3}$

= $\frac{30 - 25}{3}$

= $\frac{5}{3}$ --------------- (5)

Putting equation (5) in (4) we get,

n× $\frac{5}{3}$ = 10

=>n = 10 × $\frac{3}{5}$

=>n = 6

So putting the value of n in equation (1) we get,

a6 = 729

=>a⁶= 36

=>a = 3

And putting a = 3 in equation (5) we get,

$\frac{b}{a}$ = $\frac{5}{3}$

=>b = $\frac{5}{3}$ ×a

= $\frac{5}{3}$ × 3

= 5

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If ²⁰C_r is the co-efficient of x^r in the expansion of (1 + x)²⁰, then the value of $\sum_{r = 0}^{20} r^{2}$ ²⁰C_r is equal to

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If ²⁰C_r is the co-efficient of x^r in the expansion of (1 + x)²⁰, then the value of $\sum_{r = 0}^{20} r^{2}$ ²⁰C_r is equal to