∑ n = 1 ∞ n ! 2 n 2 + n + 1 =

∑ n = 1 ∞ n ! 2 n ( n − 1 ) + 3 n + 1 = ∑ n = 1 ∞ n ! 2 n ( n − 1 ) + ∑ n = 1 ∞ n ! 3 n + ∑ n = 1 ∞ n ! 1 2 e + 3 e + e − 1 = 6 e − 1

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Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremLogarithmic and Exponential seriesHard2 minQB

MathematicsHardsingle choice

$\sum_{n = 1}^{\infty} \frac{2 n ^{2} + n + 1}{n !} =$

Options:

Answer:

Solution:

$\sum_{n = 1}^{\infty} \frac{2 n ( n - 1 ) + 3 n + 1}{n !} = \sum_{n = 1}^{\infty} \frac{2 n ( n - 1 )}{n !} + \sum_{n = 1}^{\infty} \frac{3 n}{n !} + \sum_{n = 1}^{\infty} \frac{1}{n !}$
$2 e + 3 e + e - 1 = 6 e - 1$

Stream:JEESubject:MathematicsTopic:Binomial TheoremSubtopic:Logarithmic and Exponential series

⏱ 2mℹ️ Source: QB

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