The coefficient of \(x^{50}\) in the expansion of \((\mathrm{l}+x)^{1000}+x(\mathrm{l}+x)^{999}\)…
The coefficient of \(x^{50}\) in the expansion of \((\mathrm{l}+x)^{1000}+x(\mathrm{l}+x)^{999}\) \(+x^2(1+x)^{998}+\ldots+x^{1000}\) is
\({ }^{1000} C_{50}\)
\({ }^{999} \mathrm{C}_{50}\)
\({ }^{1000} C_{51}\)
\({ }^{1001} C_{50}\)
Solution
Expression given is,
\(\begin{aligned}
& f(x)=(l+x)^{1000}+x(l+x)^{999}+x^2(l+x)^{998} +\ldots \ldots .+x^{1000},
\end{aligned}\)
This is a Geometric progression, common ratio \(=\frac{x}{1+x}\) and number of terms 1001 .
So,
\(\begin{aligned}
f(x) & =\frac{(1+x)^{1000}\left(1-\left(\frac{x}{1+x}\right)^{1001}\right)}{\left(1-\frac{x}{1+x}\right)} \\
& =(1+x)^{1001}-x^{1001}
\end{aligned}\)
So, coefficient of \(x^{50}\)
\(\begin{aligned}
& =\text { coefficient of } x^{50} \text { in }(1+x)^{1001} \\
& ={ }^n C_{50}={ }^{1001} C_{50}
\end{aligned}\)