x^5 + 10x^4a + 40x^3a^2 + 80x^2a^3 + 80xa^4 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Using the Binomial Theorem,the expansion of $(x + y)^n$ is given by $\sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$.For $(x + 2a)^5$,we have $n=5$ and $

Vedclass · Accepted Answer

$(x + 2a)^5$

Vedclass · Answer

(C) Using the Binomial Theorem,the expansion of $(x + y)^n$ is given by $\sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$.For $(x + 2a)^5$,we have $n=5$ and $y=2a$.The expansion is:$\binom{5}{0}x^5 + \binom{5}{1}x^4(2a) + \binom{5}{2}x^3(2a)^2 + \binom{5}{3}x^2(2a)^3 + \binom{5}{4}x(2a)^4 + \binom{5}{5}(2a)^5$$= x^5 + 5x^4(2a) + 10x^3(4a^2) + 10x^2(8a^3) + 5x(16a^4) + 32a^5$$= x^5 + 10x^4a + 40x^3a^2 + 80x^2a^3 + 80xa^4 + 32a^5$.Therefore,the correct option is $C$.

$x^5 + 10x^4a + 40x^3a^2 + 80x^2a^3 + 80xa^4 + 32a^5 = $

Similar Questions

For natural numbers $m, n$,if $(1 - y)^m(1 + y)^n = 1 + a_1y + a_2y^2 + \ldots$ and $a_1 = a_2 = 10$,then $(m, n) = \_\_\_\_\_\_$.

Find $(a+b)^{4}-(a-b)^{4}$. Hence,evaluate $(\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}$. (in $\sqrt{6}$)

In the expansion of $(1 + x)^5$,the sum of the coefficients of the terms is

The sum of coefficients in the expansion of $(x + 2y + 3z)^8$ is

If $C_{j}$ stands for ${ }^{n} C_{j}$,then $\frac{C_0}{2} + \frac{C_1}{2 \cdot 2^2} + \frac{C_2}{3 \cdot 2^3} + \ldots + \frac{C_{n}}{(n+1) 2^{n+1}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module