Question

The coefficient of $x^{49}$ in the product $(x-1)(x-3)\dots (x-99)$ is

• A. $-{99}^2$
• B. $1$
• C. $-2500$
• D. none of these

Answer 1

The correct option is C $-2500$

$(x-{\alpha }_1)(x-{\alpha }_2)(x-{\alpha }_3)\dots (x-{\alpha }_n)=x^n+A_1{x}^{n-1}+A_2{x}^{n-2}+\cdots +A_n$

Where ,

$A_p=(-1{)}^p\sum ({\alpha }_1{\alpha }_2\dots {\alpha }_p);$$(p$ terms at a time$)$

Given,

$(x-1)(x-3)\dots (x-99)$ contains $50$ terms,

Coefficient of $x^{49}=A_1=(-1{)}^1\sum (\alpha )=-(1+3+\cdots +99)=-50\times 50=-2500$

$($Sum of first $n$ odd numbers $=n^2)$