JEE Advanced 2019 Paper 1 · Mathematics · Sequences and Series · hard

Find the value of Σn=1^∞(n^3/((n+2)!). (Use e≈2.71828.)

Numerical answer

Find the value of

\displaystyle\sum_{n=1}^{\infty}\frac{n^{3}}{(n+2)!}

. Give the answer correct to two decimal places. (Use

e\approx2.71828

Show answer and solution

Answer: 0.85

Express

n^3

in the factorial-friendly basis

(n+2)(n+1)n

(n+2)(n+1)

(n+2)

and

1

. One finds

n^3=(n+2)(n+1)n-3(n+2)(n+1)+7(n+2)-8

. Dividing by

(n+2)!

, the general term becomes

\frac1{(n-1)!}-\frac3{n!}+\frac7{(n+1)!}-\frac8{(n+2)!}

. The sums over

n\ge1

are

\sum\frac1{(n-1)!}=e

\sum\frac1{n!}=e-1

\sum\frac1{(n+1)!}=e-2

and

\sum\frac1{(n+2)!}=e-\frac52

. Therefore the total is

e-3(e-1)+7(e-2)-8\left(e-\frac52\right)=9-3e\approx9-8.15485=0.84515

, which is

0.85

to two decimal places. A numerical partial sum of the first few terms confirms the value.

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