COMEDK · Maths · 5. Sequences and Series
The value of \(\frac{1}{2 !}+\frac{2}{3 !}+\ldots+\frac{99}{100 !}\) is equal to
- A \(\frac{100 !-1}{100 !}\)
- B \(\frac{100 !+1}{100 !}\)
- C \(\frac{999 !-1}{999 !}\)
- D \(\frac{999 !+1}{999 !}\)
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…
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