Question

The coefficient of $x^{50}$ in the expansion of $(1+x{)}^{1000}+2x(1+x{)}^{999}+3x^2(1+x{)}^{989}+.....+1001x^{1000}$ is

• A. ${}^{1000}{C}_{50}$
• B. ${}^{1001}{C}_{50}$
• C. ${}^{1002}{C}_{50}$
• D. $2^{1001}$

Answer 1

The correct option is C ${}^{1002}{C}_{50}$

Given

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

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

Now,

$S[1-\frac{x}{x+1}]=\underset{S^1}{\underset{}{(1+x{)}^{1000}+[x(1+4{)}^{999}+x^2(1+x{)}^{998}+...+x^{1000})}}-\frac{1001x^{1001}}{1+x}$

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

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

$\frac{S^1}{1+x}=x(1+x{)}^{999}-\frac{x^{1001}}{1+x}\Rightarrow s^1$

$=x(1+x{)}^{1000}-x^{1001}$

So we get

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

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

Now coff. of $x^{50}$

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

$={}^{1002}{C}_{50}$