Question

The value of ${lim}_{x\to 1}\frac{x+x^2+x^3+......x^n}{x-1}$ is

• A. $\frac{n+1}{2}$
• B. $\frac{n(n+1)}{2}$
• C. $\frac{n(n-1)}{2}$
• D. $0$

Answer 1

Answer: (B)

$\frac{n(n+1)}{2}$

$\underset{x\to 1}{lt}\frac{x+x^2+x^3+....x^n}{x-1}$

Using GP formula,

$=\frac{x\times \frac{x^n-1}{x-1}-n}{x-1}$

Simplifying,

$=\frac{x^{n+1}-x-nx+1}{(x-1{)}^2}$

Now it is $\frac{0}{0}$ form LH rule,

$=\frac{(n+1)x^n-1-n}{2(x-1)}$ (Still $\frac{0}{0}$ form)

LH rule again

$=\frac{1}{2}\left[\right(n+1\left)n{x}^{n-1}\right]$

Put $x=1$

$\therefore B=\frac{n(n+1)}{2}$