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  • The coefficient of  in the expression <img alt="(1+x)^{1000}+2 x(1+x)^{999}+" src="https://en

The coefficient of x^{100} in the expression (1+x)^{1000}+2 x(1+x)^{999}+3 x^2(1+x)^{998}+\ldots+1001 x^{1000} \text { is }

Option: 1

{ }^{1000} C_{100}


Option: 2

{ }^{1001} C_{100}


Option: 3

{ }^{1002} C_{100}


Option: 4

None of these


Answers (1)

best_answer

The given series is A.G.P.. Lets first find its sum.

Writing t for 1 + x, let

\begin{aligned} & S=t^{1000}+2 x t^{999}+3 x^2 t^{998}+\ldots+1001 x^{1000} \\ \therefore \quad & (x / t) \cdot S=x t^{999}+2 x^2 t^{998}+\ldots+1001 x^{1001} / t \end{aligned}Subtracting, we get

\begin{aligned} & S(1-x / t)=t^{1000}+x t^{999}+x^2 t^{998}+\ldots+x^{1000}-1001 x^{1001 / t} \\ & =\frac{t^{1000}\left[1-(x / t)^{1001}\right]}{1-x / t}-\frac{1001 x^{1001}}{t} \end{aligned}\text { or } S=(1+x)^{1002}-x^{1001}(1+x)-1001 x^{1001}

(Putting t = 1 + x and simplifying)

\therefore Coefficient of x100 in the expansion = coefficient of x100 in the expansion of (1+x)^{1002}={ }^{1002} C_{100}

Posted by

himanshu.meshram

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