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  • The value of coefficient of  in the expansion of <img alt="(1+x)^{1000}+2 x(1+x)^{999}+3 x^2(1+

The value of coefficient of x^{50} in the expansion of (1+x)^{1000}+2 x(1+x)^{999}+3 x^2(1+x)^{998}+\cdots+1001 x^{1000} is 

Option: 1

\left(\begin{array}{c} 1002 \\ 50 \end{array}\right)


Option: 2

\left(\begin{array}{c} 1002 \\ 52 \end{array}\right)


Option: 3

\left(\begin{array}{c} 1004 \\ 50 \end{array}\right)


Option: 4

None


Answers (1)

best_answer

Let \\S=(1+x)^{1000}+2 x(1+x)^{999}+. .1000 x^{999}(1+x)+1001 x^{1000} \ldots \ldots(1) \times \frac{x}{(1+x)}

So we get \frac{x \cdot S}{(1+x)}=x(1+x)^{999}+2 x^2(1+x)^{998}+\ldots \ldots \ldots+1000 x^{1000}+\frac{1001 x^{1001}}{1+x} \ldots

So we get S\left[1-\frac{x}{1+x}\right]=(1+x)^{1000}+\underbrace{\left[x(1+x)^{999}+x^2(1+x)^{998}+\ldots \ldots+x^{1000}\right]}_{S^{\prime}}-\frac{1001 x^{1001}}{1+x}Now S^{\prime}=x(1+x)^{999}+x^2(1+x)^{998}+\ldots \ldots \ldots \ldots+x^{1000} \ldots \ldots \ldots(1) \times \frac{x}{1+x}

So       \frac{S^{\prime} \cdot x}{1+x}=x^2(1+x)^{998}+\ldots \ldots \ldots+\frac{x^{1001}}{1+x} \ldots \ldots .(2)

So we get\frac{S^{\prime}}{1+x}=x(1+x)^{999}-\frac{x^{1001}}{1+x} \Rightarrow S^{\prime}=x(1+x)^{1000}-x^{1001}

So we get \frac{S}{1+x}=(1+x)^{1000}+x(1+x)^{1000}-x^{1001}-\frac{1001 x^{1001}}{1+x}

So we get  S=(1+x)^{1002}-x^{1001}(1+x)-1001 x^{1001}

Now Coefficient of x^{50} in (1+x)^{1002}-x^{1001}(1+x)-1001 x^{1001} is =\left(\begin{array}{c} 1002 \\ 50 \end{array}\right)

Posted by

Ritika Harsh

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