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JEE Mains · Maths · STD 11 - 8. sequence and series

શ્રેઢી \(\sum_{n=1}^{\infty} \frac{n^{2}+6 n+10}{(2 n+1) !}\) નો સરવાળો ..... થાય.

  1. A \(\frac{41}{8} e +\frac{19}{8} e ^{-1}-10\)
  2. B \(\frac{41}{8} e -\frac{19}{8} e ^{-1}-10\)
  3. C \(\frac{41}{8} e +\frac{19}{8} e ^{-1}+10\)
  4. D \(-\frac{41}{8} e +\frac{19}{8} e ^{-1}-10\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{41}{8} e -\frac{19}{8} e ^{-1}-10\)

Step-by-step Solution

Detailed explanation

\(T _{ n }=\frac{ n ^{2}+6 n +10}{(2 n +1) !}=\frac{4 n ^{2}+24 n +40}{4 \cdot(2 n +1) !}\) \(=\frac{(2 n+1)^{2}+20 n+39}{4 \cdot(2 n+1) !}\) \(=\frac{(2 n+1)^{2}+(2 n+1) \cdot 10+29}{4(2 n+1) !}\)…
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