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STEP III 1997 question 1 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1997 question 1 solution

(i)

$\displaystyle (x^2 + 5x + 4)e^x = \sum_{n=0}^{\infty} \frac{x^(n+2)}{n!} + 5 \sum_{n=0}^{\infty} \frac{x^(n+1)}{n!} + 4 \sum_{n=0}^{\infty} \frac{x^n}{n!}$

$\displaystyle = \sum_{n=2}^{\infty} \frac{x^n}{(n-2)!} + 5 \sum_{n=1}^{\infty} \frac{x^n}{(n-1)!} + 4 \sum_{n=0}^{\infty} \frac{x^n}{n!}$

$\displaystyle = 4 + 9x + \sum_{n=2}^{\infty} \left[\left(\frac{1}{(n-1)!} + \frac{5}{(n-1)!} + \frac{4}{n!}\right) x^n\right]$

$\displaystyle = 4 + 9x + \sum_{n=2}^{\infty} \left[\frac{n(n-1) + 5n + 4}{n!} x^n\right]$

$\displaystyle = 4 + 9x + \sum_{n=2}^{\infty} \frac{(n+2)^2}{n!} x^n$

Now plug in x=1 to get:

$\displaystyle 10e = 4 + \frac{3^2}{1!} + \frac{4^2}{2!} + \frac{5^2}{3^2} + \cdots$ , as required.

(ii)

$\displaystyle 10e = \sum_{n=0}^{\infty} \frac{(n+2)^2}{n!}$

$\displaystyle = \sum_{n=0}^{\infty} \frac{n^2 + 2n + 4}{n!}$

$\displaystyle = \sum_{n=0}^{\infty} \frac{(n+1)^2}{n!} + \sum_{n=0}^{\infty} \frac{2n+3}{n!}$

$\displaystyle = \sum_{n=0}^{\infty} \frac{(n+1)^2}{n!} + 3 + \sum_{n=1}^{\infty} \frac{2n+3}{n!}$

$\displaystyle = \sum_{n=0}^{\infty} \frac{(n+1)^2}{n!} + 3 + 2 \sum_{n=1}^{\infty} \frac{2}{(n-1)!} + 3 \sum_{n=1}^{\infty} \frac{1}{n!}$

$\displaystyle = \sum_{n=0}^{\infty} \frac{(n+1)^2}{n!} + 3 + 2e + 3(e-1)$

So

$\displaystyle 10e = \sum_{n=0}^{\infty} \frac{(n+1)^2}{n!} + 3 + 5e - 3$

$\displaystyle \Longrightarrow 5e = \sum_{n=0}^{\infty} \frac{(n+1)^2}{n!}$ , as required.

(iii)

$\displaystyle \sum_{n=0}^{\infty} \frac{(n+1)^3}{n!} = \sum_{n=0}^{\infty} \frac{(n+1)(n+1)^2}{/n!}$

$\displaystyle = \sum_{n=0}^{\infty} \frac{n(n+1)^2}{n!} + \sum_{n=0}^{\infty} \frac{(n+1)^2}{n!}$

= 10e + 5e

= 15e

Solution by dvs

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