Mathematics - Binomial Theorem Question with Solution | TestHub

Let $S=(1+x{)}^{1000}+2x(1+x{)}^{999}+3x^2(1+x{)}^{998}+\dots +1001x^{1000}$.

Multiply $S$ by $\frac{x}{1+x}$:

$$\frac{x}{1+x}S=x(1+x{)}^{999}+2x^2(1+x{)}^{998}+\dots +1000x^{1000}+\frac{1001x^{1001}}{1+x}$$

Subtracting the second equation from the first gives:

$$\begin{gathered} \text{displaylines}\left(1-\frac{x}{1+x}\right)S= \\ (1+x{)}^{1000}+x(1+x{)}^{999}+x^2(1+x{)}^{998}+\dots +x^{1000}-\frac{1001x^{1001}}{1+x} \end{gathered}$$

$$\frac{1}{1+x}S=(1+x{)}^{1000}+x(1+x{)}^{999}+x^2(1+x{)}^{998}+\dots +x^{1000}-\frac{1001x^{1001}}{1+x}$$

$$S=(1+x{)}^{1001}+x(1+x{)}^{1000}+x^2(1+x{)}^{999}+\dots +x^{1000}(1+x)-1001x^{1001}$$

The terms $(1+x{)}^{1001}+x(1+x{)}^{1000}+x^2(1+x{)}^{999}+\dots +x^{1000}(1+x)$ form a geometric progression.

The sum of this geometric progression is:

$$\frac{(1+x{)}^{1001}\left[{\left(\frac{x}{1+x}\right)}^{1001}-1\right]}{\frac{x}{1+x}-1}=\frac{(1+x{)}^{1001}\left[\frac{x^{1001}}{(1+x{)}^{1001}}-1\right]}{\frac{x-(1+x)}{1+x}}$$

$$=\frac{(1+x{)}^{1001}\left[\frac{x^{1001}-(1+x{)}^{1001}}{(1+x{)}^{1001}}\right]}{\frac{-1}{1+x}}$$

$$=\frac{x^{1001}-(1+x{)}^{1001}}{\frac{-1}{1+x}}=-(1+x)(x^{1001}-(1+x{)}^{1001})$$

$$=(1+x)(1+x{)}^{1001}-(1+x)x^{1001}=(1+x{)}^{1002}-x^{1001}(1+x)$$

So, $S=(1+x{)}^{1002}-x^{1001}(1+x)-1001x^{1001}$.

$S=(1+x{)}^{1002}-x^{1001}-x^{1002}-1001x^{1001}$.

$S=(1+x{)}^{1002}-x^{1002}-1002x^{1001}$.

We need to find the coefficient of $x^{50}$ in $S$.

The coefficient of $x^{50}$ in $S$ is the coefficient of $x^{50}$ in $(1+x{)}^{1002}-x^{1002}-1002x^{1001}$.

Since $50<1001$ and $50<1002$, the terms $-x^{1002}$ and $-1002x^{1001}$ do not contribute to the coefficient of $x^{50}$.

Therefore, the coefficient of $x^{50}$ in $S$ is the coefficient of $x^{50}$ in $(1+x{)}^{1002}$, which is ${}^{1002}{C}_{50}$.