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The sum of the series \(1 + \frac{{{2^3}}}{{2!}} + \frac{{{3^3}}}{{3!}} + \frac{{{4^3}}}{{4!}} + \cdot \cdot \cdot \) is
1. 5e - 1
2. 5e - 3
3. 4e
4. None of these

1 Answer

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Correct Answer - Option 4 : None of these

Concept:

Expansion of exponential function.

\(e^x = \displaystyle\sum_{n = 0}^\infty \dfrac{x^n}{n!}\)

\(=\dfrac{1}{1}+\dfrac{x}{1}+ \dfrac{x^2}{2!} + \dfrac{x^3}{3!}+...\)

From the above formula value of e1 and e-1 are. 

\(e^1 = \displaystyle\sum_{n=0}^\infty \dfrac{1}{n!}\)

\(=\dfrac{1}{1} + \dfrac{1}{1!} + \dfrac{1^2}{2!} + \dfrac{1^3}{3!} + ...\)

\(e^{-1} = \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n}{n!}\)

\(=\dfrac{1}{1} - \dfrac{1}{1} + \dfrac{1}{2!} - \dfrac{1}{6} + ...\)

Calculation:

Given expansion is:

\(1 + \dfrac{2^3}{2!} + \dfrac{3^3}{3!} + \dfrac{4^3}{4!} + ...\)

we can verify the options and conclude this

Option (A) 5e - 1

\(5 \left[1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + ... \right]-1\)

\(5 + 5 + \dfrac{5}{2} + \dfrac{5}{6} + ...-1\)

\(9 + \dfrac{5}{2} + \dfrac{5}{6} + ...\)

not equal to the given series.

option (B) 5e - 3

\(5\left[1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + ...\right] - 3\)

\(7 + \dfrac{5}{2} + \dfrac{5}{6} + ...\)

not equal to given series.

Option (c) 4e

\(4 \left[1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + ...\right]\)

\(4 + \dfrac{4}{1!} + \dfrac{4}{4!} + \dfrac{4}{3!} + ...\)

not equal to given series

The answer is option D.

None of the above expansions are matching to given series.

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