NIMCET 2023 — Mathematics PYQ

Sol. (c)
Let $S=(1+x{)}^{1000}+2x(1+x{)}^{999}+3x^2(1+x{)}^{998}+\dots +1000x^{999}(1+x)+1001x^{1000}$
Above is A.G.P. of common ratio $r=\frac{x}{1+x}$
$\therefore \left[\frac{x}{(1+x)}\right]S=x(1+x{)}^{999}+2x^2(1+x{)}^{998}+\dots +1000\cdot x^{1000}+\frac{1001x^{1001}}{1+x}$
Subtracting, $\left(1-\frac{x}{1+x}\right)S=(1+x{)}^{1000}+x(1+x{)}^{999}+x^2(1+x{)}^{998}+\dots +x^{1000}-\frac{1001x^{1001}}{1+x}$
Or, $S=(1+x{)}^{1001}+x(1+x{)}^{1000}+x^2(1+x{)}^{999}+\dots +x^{1000}(1+x)-1001x^{1001}$
$$=\frac{(1+x{)}^{1001}\left[1-{\left(x-(1+x)\right)}^{1001}\right]}{1-x}-1001x^{1001}$$
Sum G.P. $(1+x{)}^{1002}\left[1-{\left(\frac{x}{(1+x)}\right)}^{1001}\right]-1001x^{1001}$
$$<br>=(1+x{)}^{1002}-x^{1001}(1+x)-1001x^{1001}$$
$$=(1+x{)}^{1002}-x^{1002}-1002x^{1001}\dots (i)$$
Now the coefficients of $x^{50}$ on the R.H.S. of (i)
$$<br>={}^{1002}{C}_{50}$$