WBJEE · Maths · Sequences and Series
The value of the infinite series \(\frac{1^{2}+2^{2}}{3 !}\) \(+\frac{1^{2}+2^{2}+3^{2}}{4 !}+\frac{1^{2}+2^{2}+3^{2}+4^{2}}{5 !}+\ldots\)
- A \(e\)
- B \(5 e\)
- C \(\frac{5 e}{6}-\frac{1}{2}\)
- D \(\frac{5 e}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{5 e}{6}-\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Given infinite series is \(\frac{1^{2}+2^{2}}{3 !}+\frac{1^{2}+2^{2}+3^{2}}{4 !}\) \(\begin{aligned} \frac{1^{2}+2^{2}}{3 !} &+\frac{1^{2}+2^{2}+3^{2}}{4 !} \\ &+\frac{1^{2}+2^{2}+3^{2}+4^{2}}{5 !}+\ldots \end{aligned}\) nth term,…
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