∑ k = 1 ∞ k ! 1 ( ∑ n = 1 k 2 n − 1 ) =

∑ k = 1 ∞ k ! 1 ( 1 + 2 + 2 2 + …… + 2 k − 1 ) ∑ k = 1 ∞ k ! 2 k − 1 = e 2 − e

Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremLogarithmic and Exponential seriesHard2 minQB

MathematicsHardsingle choice

$\sum_{k = 1}^{\infty} \frac{1}{k !} (\sum_{n = 1}^{k} 2^{n - 1}) =$

Answer:

Solution:

$\sum_{k = 1}^{\infty} \frac{1}{k !} (1 + 2 + 2^{2} + \dots\dots + 2^{k - 1})$

$\sum_{k = 1}^{\infty} \frac{2 ^{k} - 1}{k !} = e^{2} - e$

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

⏱ 2mℹ️ Source: QB

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