The value of 2 1 ! + 2 + 4 2 ! + 2 + 4 + 6 3 ! + … . is

Here, T n = n ( n + 1 ) n ! = n - 1 + 2 ( n - 1 ) ! = 1 ( n - 2 ) ! + 2 ( n - 1 ) ! ∴ S = ∑ n = 1 ∞ T n = ∑ n = 1 ∞ 1 ( n - 2 ) ! + 2 ∑ n = 1 ∞ 1 ( n - 1 ) ! = e + 2 e = 3 e

Mathematics - Basic maths Question with Solution | TestHub

MathematicsBasic mathsLogarithmMedium2 minPYQ_2017

MathematicsMediumsingle choice

The value of $\frac{2}{1!} + \frac{2 + 4}{2!} + \frac{2 + 4 + 6}{3!} + \dots .$ is

Options:

Answer:

Solution:

Here,

$T_{n} = \frac{n (n + 1)}{n!} = \frac{n - 1 + 2}{(n - 1)!} = \frac{1}{(n - 2)!} + \frac{2}{(n - 1)!}$

$∴ S = \sum_{n = 1}^{\infty} T_{n} = \sum_{n = 1}^{\infty} \frac{1}{(n - 2)!} + 2 \sum_{n = 1}^{\infty} \frac{1}{(n - 1)!}$

$= e + 2 e = 3 e$

Stream:JEESubject:MathematicsTopic:Basic mathsSubtopic:Logarithm

⏱ 2mℹ️ Source: PYQ_2017

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