Question

Simplify the expression
${(1+x)}^{1000}+x(1+x{)}^{999}+x^2(1+x{)}^{998}+\cdots +x^{1000}$ and find the coefficient of $x^{50}$

Answer 1

Clearly, the given series is a geometric series in which
$a=(1+x{)}^{1000},r=\frac{x(1+x{)}^{999}}{(1+x{)}^{1000}}=\frac{x}{(1+x)}andn=1001\therefore \text{Given sum}=\frac{a(1-r^n)}{(1-r)}\frac{(1+x{)}^{1000}\times \{1-{\left(\frac{x}{1+x}\right)}^{1001}\}}{(1-\frac{x}{1+x})}=(1+x{)}^{1001}-x^{1001}1{+}^{1001}{C}_{1}x{+}^{1001}{C}_2{x}^2+{\cdots }^{1001}{C}_{1000}{x}^{1000}=1{+}^{1001}{C}_{1}x{+}^{1001}{C}_2{x}^2+\cdots {+}^{1001}{C}_1{x}^{1000}\therefore \text{coefficient of}{x}^{50}\text{in the above expansion}{=}^{1001}{C}_{50}=\frac{\left(1001\right)!}{\left(50\right)!Q.(1001-50)!}=\frac{\left(1001\right)!}{\left(50\right)!.\left(951\right)!}$