1+(2^3)/(2!)+(3^3)/(3!)+(4^3)/(4!)+… ∞ equals

Correct answer: (A) 5 e. Here, T_n=(n^3)/(n!)=(n^2-1)/((n-1)!)+1/((n-1)!) =(n+1)/((n-2)!)+1/((n-1)!) =1/((n-3)!)+3/((n-2)!)+1/((n-1)!) ∴ Σ T_n=Σ (1/((n-3)!)+3/((n-2)!)+1/((n-1)!) ) =e+3 e+e=5 e

Q108 / 170 · Manipal (MET) 2016 · Mathematics

$1+\frac{2^3}{2!}+\frac{3^3}{3!}+\frac{4^3}{4!}+\ldots \infty$ equals

Correct answer
A
$5 e$
B
$4 e$
C
$3 e$
D
$2 e$

Solution

Here, $T_n=\frac{n^3}{n!}=\frac{n^2-1}{(n-1)!}+\frac{1}{(n-1)!}$ $\begin{aligned} & =\frac{n+1}{(n-2)!}+\frac{1}{(n-1)!} \\ & =\frac{1}{(n-3)!}+\frac{3}{(n-2)!}+\frac{1}{(n-1)!}\end{aligned}$ $\therefore \Sigma T_n=\Sigma\left(\frac{1}{(n-3)!}+\frac{3}{(n-2)!}+\frac{1}{(n-1)!}\right)$ $=e+3 e+e=5 e$

Hence, the correct answer is option: 'A'.

$1+\frac{2^3}{2!}+\frac{3^3}{3!}+\frac{4^3}{4!}+\ldots \infty$ equals

Solution

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