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.