- Home
- Standard 11
- Mathematics
The coefficient of $x^{50}$ in the binomial expansion of ${\left( {1 + x} \right)^{1000}} + x{\left( {1 + x} \right)^{999}} + {x^2}{\left( {1 + x} \right)^{998}} + ..... + {x^{1000}}$ is
$\frac{{\left( {1000} \right)!}}{{\left( {50} \right)!\left( {950} \right)!}}$
$\frac{{\left( {1000} \right)!}}{{\left( {49} \right)!\left( {951} \right)!}}$
$\frac{{\left( {1001} \right)!}}{{\left( {51} \right)!\left( {950} \right)!}}$
$\frac{{\left( {1001} \right)!}}{{\left( {50} \right)!\left( {951} \right)!}}$
Solution
Let given expansion be
$\mathrm{S}=(1+x)^{1000}+x(1+x)^{999}+x^{2}$
$(1+x)^{998}+\ldots+\ldots+x^{1006}$
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 $\frac{x}{t}$
$S=\frac{t^{1000}\left[1-\left(\frac{x}{t}\right)^{1001}\right]}{1-\frac{x}{t}}$
${ = \frac{{{{(1 + x)}^{1000}}\left[ {1 – {{\left( {\frac{x}{{1 + x}}} \right)}^{1001}}} \right]}}{{1 – \frac{x}{{1 + x}}}}}$
$ = \frac{{{{(1 + x)}^{1001}}\left[ {{{(1 + x)}^{1001}} – {x^{1001}}} \right]}}{{{{(1 + x)}^{1001}}}}$
$=\left[(1+x)^{100 t}-x^{1001}\right]$
Now coeff of $x^{50}$ in above expansion is equal to coeff of $x^{50}$ in $(1+x)^{1001}$ which is
$^{1001} C_{50}=\frac{-(1001) !}{50 !(951) !}$
