TS EAMCET · Maths · Sequences and Series
\(\frac{1}{2 !}+\frac{1+2}{3 !}+\frac{1+2+3}{4 !}+\ldots\) is equal to :
- A \(\frac{e}{2}\)
- B \(\frac{e}{3}\)
- C \(\frac{e}{4}\)
- D \(\frac{e}{5}\)
Answer & Solution
Correct Answer
(A) \(\frac{e}{2}\)
Step-by-step Solution
Detailed explanation
We have \(n\)th term \(T_n=\frac{1+2+3+\ldots+n}{(n+1) !}\) \(=\frac{n(n+1)}{2 \cdot(n+1) !}=\frac{1}{2(n-1) !}\) Put \(n=1,2,3, \ldots\)…
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