The coefficient of $x^{50}$ in the binomial expansion of $(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\ldots$…
The coefficient of $x^{50}$ in the binomial expansion of $(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\ldots$ $+x^{1000}$ is:
$\frac{(1000) !}{(50)(! 95 \varphi !}$
$\frac{(1000) !}{(49)(! 95) !}$
$\frac{(1001) !}{(51)(! 95 \phi !}$
$\frac{(1001) !}{(50)(! 95) !}$
Solution
Let given expansion be
$
\begin{aligned}
&\mathrm{S}=(1+x)^{1000}+x(1+x)^{999}+x^2 \\
&(1+x)^{998}+\ldots+\ldots+x^{1000}
\end{aligned}
$
Put $1+x=t$
$
\mathrm{S}=t^{1000}+x t^{999}+x^2(t)^{998}+\ldots+x^{1000}
$
This is a G.P with common ratio $\frac{x}{t}$
$
\begin{aligned}
\mathrm{S} &=\frac{t^{1000}\left[1-\left(\frac{x}{t}\right)^{1001}\right]}{1-\frac{x}{t}} \\
&=\frac{(1+x)^{1000}\left[1-\left(\frac{x}{1+x}\right)^{1001}\right]}{1-\frac{x}{1+x}} \\
=& \frac{(1+x)^{1001}\left[(1+x)^{1001}-x^{1001}\right]}{(1+x)^{1001}}
\end{aligned}
$
$
=\left[(1+x)^{1001}-x^{1001}\right]
$
Now coeff of $x^{50}$ in above expansion is equal to coeff of $x^{50}$ in $(1+x)^{1001}$ which is ${ }^{1001} C_{50}=\frac{(1001) !}{50 !(951) !}$