Q100 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics

$\lim _{x \rightarrow 1}\left(\frac{x+x^2+x^3+\ldots+x^n-n}{x-1}\right)=$

  • Correct answer
    A
    $\frac{n(n+1)}{2}$
  • B
    $\frac{n+1}{2}$
  • C
    $\frac{2}{n}$
  • D
    $n$

Solution

$\begin{aligned} & \text { } \lim _{x \rightarrow 1}\left(\frac{x+x^2+x^3 \ldots . x^4-n}{(x-1)}\right) \\ & =\lim _{x \rightarrow 1}\left(\frac{(x-1)+\left(x^2-1\right)+\left(x^3-1\right) \ldots+\left(x^n-1\right)}{(x-1)}\right) \\ & =\lim _{x \rightarrow 1}\left(1+2 x+3 x^2+4 x^3 \ldots . . . n x^{n-1}\right) \\ & =1+2+3+4+\ldots . . h=\frac{n(n+1)}{2} .\end{aligned}$

Hence, the correct answer is option: 'A'.

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