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COMEDK · Maths · 5. Sequences and Series

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

  1. A \(\frac{100 !-1}{100 !}\)
  2. B \(\frac{100 !+1}{100 !}\)
  3. C \(\frac{999 !-1}{999 !}\)
  4. D \(\frac{999 !+1}{999 !}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{100 !-1}{100 !}\)

Step-by-step Solution

Detailed explanation

\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…