Question

Find the coefficient of $x^{50}$ in the expression
$(1+x{)}^{1000}+2x(1+x{)}^{999}+3x^2(1+x{)}^{998}+1000x^{1000}$

Answer 1

Let S = $(1+x{)}^{1000}+2x(1+x{)}^{999}+3x^2(1+x{)}^{998}+...+$
$1000x^{999}(1+x)+1001x^{1000}$
Above is A.G.S. of common ratio r = $\frac{x}{1+x}$
$\therefore \left[x/\right(1+x\left)\right]S=x(1+x{)}^{999}+2x^2(1+x{)}^{998}+$
$.....+1000\cdot x^{1000}+\frac{1001x^{1001}}{1+x}$
Subtracting,
$(1-\frac{x}{1+x})S=(1+x{)}^{1000}+x(1+x{)}^{999}$
$+x^2(1+x{)}^{998}+....+x^{1000}-\frac{1001x^{1001}}{1+x}$
or S = $(1+x{)}^{1001}+x(1+x{)}^{1000}+x^2(1+x{)}^{999}+$
$......+x^{1000}(1+x)-1001x^{1001}$
$=\frac{(1+x{)}^{1001}[1-\left(x\right(1+x\left){)}^{10001}\right]}{1-x}-1001x^{1001}$
Sum G.P.
$(1+x{)}^{1002}[1-\left(x/\right(1+x\left){)}^{1001}\right]-1001x^{1001}$
$=(1+x{)}^{1002}-x^{1001}(1+x)-1001x^{1001}$
$=(1+x{)}^{1002}-x^{1002}-1002x^{1001}$ ....(1)
Now the coefficients of $x^{50}$ on the R.H.S. of (1)
$={}^{1002}{C}_{50}$