Question

The coefficient of $x$ in the expansion of $(1+x)(1+2x)(1+3x)....(1+100x)$ is

• A. $5050$
• B. $10100$
• C. $5151$
• D. $4950$

Answer 1

The correct option is A $5050$

$(1+x)(1+2x)(1+3x)......(1+100x)$

We are required to find the coefficient of ${}^{\prime }{x}^{\prime }$ in the above expansion.

Now, first multiply the first two terms

$(1+x)(1+2x)=1+x+2x+{2x}^2=1+(1+2)x+2x^2$

Similarly multiply the first three terms

$(1+x)(1+2x)(1+3x)=1+(1+2+3)x+{11x}^2+{6x}^3$

So in the expansion of $(1+x)(1+2x)(1+3x)......(1+100x)$

coefficient of $x$ should be $(1+2+3+........+100)$

We know that the sum of the first ${}^{\prime }{n}^{\prime }$ natural numbers is given by the expression $\frac{n(n+1)}{2}$

$\therefore$ Coefficient of $x$ in the above expansion is $=\frac{100\times 101}{2}=5050$.

Correct answer : Option A.