Q95 / 180 · COMEDK 2023 (Memory Based) · Mathematics

The value of $\frac{1}{2 !}+\frac{2}{3 !}+\ldots+\frac{99}{100 !}$ is equal to

  • Correct answer
    A
    $\frac{100 !-1}{100 !}$
  • B
    $\frac{100 !+1}{100 !}$
  • C
    $\frac{999 !-1}{999 !}$
  • D
    $\frac{999 !+1}{999 !}$

Solution

$\begin{aligned} & \text { Given, } \frac{1}{2 !}+\frac{2}{3 !}+\ldots+\frac{99}{100 !} \\ & =\frac{2-1}{2 !}+\frac{3-1}{3 !}+\frac{4-1}{4 !}+\ldots+\frac{100-1}{100 !} \\ & =\left(\frac{1}{1 !}-\frac{1}{2 !}\right)+\left(\frac{1}{2 !}-\frac{1}{3 !}\right)+\left(\frac{1}{3 !}-\frac{1}{4 !}\right)+\ldots+\left(\frac{1}{99 !}-\frac{1}{100 !}\right) \\ & =1-\frac{1}{100 !}=\frac{100 !-1}{100 !}\end{aligned}$

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

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