Question

Let $f\left(x\right)+1=i-j+\sum _{n=i}^j{x}^n$, where $0<i<j$. If $\underset{x\to 1}{lim}\frac{f\left(x\right)}{x-1}=50(i+j)$, then the value of $(j-i)$ is

• A. $49$
• B. $51$
• C. $99$
• D. $101$

Answer 1

The correct option is C $99$

$f\left(x\right)=i-j-1+\sum _{n=i}^j{x}^n$

$\underset{x\to 1}{lim}\frac{f\left(x\right)}{x-1}=\underset{x\to 1}{lim}\frac{i-j-1}{x-1}+\frac{1}{x-1}\sum _{n=i}^j{x}^n$
$=\underset{x\to 1}{lim}\frac{x^i+x^{i+1}+x^{i+2}+...+x^j+i-j-1}{x-1}=\underset{x\to 1}{lim}(\frac{x^i-1}{x-1}+\frac{x^{i+1}-1}{x-1}+\frac{x^{i+2}-1}{x-1}+...+\frac{x^j-1}{x-1})=i+(i+1)+(i+2)+...+j=\frac{j-i+1}{2}[2i+(j-i)\times 1],[\because a=i,n=j-i+1]=\left(\frac{j-i+1}{2}\right)(i+j)\Rightarrow \left(\frac{j-i+1}{2}\right)(i+j)=50(i+j)\Rightarrow j-i=99$