Question

Find the coefficient of $x^{50}$ in the polynomials after parenthesis have been removed and like terms have been collected in the expansion
${(1+x)}^{1000}+x{(1+x)}^{999}+x^2{(1+x)}^{998}+....+x^{1000}$

Answer 1

Given expansion is

${(1+x)}^{1000}+2x{(1+x)}^{999}+3x^2{(1+x)}^{998}+3x^3{(1+x)}^{997}+....+1001x^{1000}$

Let $S$ denotes the sum of the polynomial.

$S={(1+x)}^{1000}+2x{(1+x)}^{999}+3x^2{(1+x)}^{998}+3x^3{(1+x)}^{997}+....+1001x^{1000}$ .....(1)
Multiplying by $\frac{x}{1+x}$ in eqn (1), we get

$\frac{x}{(1+x)}S=x{(1+x)}^{999}+2x^2{(1+x)}^{998}+3x^3{(1+x)}^{997}+...+1000x^{1000}+\frac{1001}{(1+x)}{x}^{1001}$ ....(2)
Subtracting eqn(2) from eqn(1), we get
$(1-\frac{x}{1+x})S=(1+x{)}^{1000}+x{(1+x)}^{999}+x^2{(1+x)}^{998}+x^3{(1+x)}^{997}+...+x^{1000}-\frac{1001}{(1+x)}{x}^{1001}$
$\Rightarrow \left(\frac{1}{1+x}\right)S=(1+x{)}^{1000}+x{(1+x)}^{999}+x^2{(1+x)}^{998}+x^3{(1+x)}^{997}+....+x^{1000}-\frac{1001}{(1+x)}{x}^{1001}$
$\Rightarrow S=\left[\right(1+x{)}^{1001}+x{(1+x)}^{1000}+x^2{(1+x)}^{999}+x^3{(1+x)}^{998}+....+x^{1000}(1+x)]-1001x^{1001}$

The sequence in square bracket forms a G.P.

Here, first term $a=(1+x{)}^{1001}$ and common ratio $r=\frac{x}{1+x}$.

So, using the formula for sum of n terms of G.P. $S_n=\frac{a(1-r^n)}{1-r}$

$\Rightarrow S=\frac{(1+x{)}^{1001}[1-{\left(\frac{x}{1+x}\right)}^{1001}]}{1-\frac{x}{1+x}}-1001x^{1001}$
$\Rightarrow S=\frac{(1+x{)}^{1001}[1-{\left(\frac{x}{1+x}\right)}^{1001}]}{\frac{1}{1+x}}-1001x^{1001}$
$\Rightarrow S=(1+x{)}^{1002}[1-{\left(\frac{x}{1+x}\right)}^{1001}]-1001x^{1001}$
$\Rightarrow S=(1+x{)}^{1002}-(1+x)x^{1001}-1001x^{1001}$

$\Rightarrow S=(1+x{)}^{1002}-1002x^{1001}-x^{1002}$
Since, the last two terms in the above sum cannot have $x^{50}$ and only the first term can have $x^{50}$

Now, the coefficient of $x^r$ in the expansion of $(1+x{)}^n$ is ${}^n{C}_r$
So, the coefficient of $x^{50}$ in $(1+x{)}^{1002}$ is ${}^{1002}{C}_{50}$