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

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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 = $

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