Logging the problem

We have:

$a_{n}=\frac{(log3)^n}{n!}\sum_{1}^{n}r^2{n\choose r}$

Also: $\sum_{1}^{n}r^2{n\choose r} = n(n+1)2^{n-2}$ Proving/Verifying this fairly trivial result is left as an exercise to the reader(just love doing that :D).

Thus: $a_{n}=\frac{(log3)^n}{n!}n(n+1)2^{n-2}=\frac{(2log3)^n*n(n+1)}{4n!}=\frac{n(n+1)(log9)^n}{4n!}$

The $n!$ in the denominator motivates us to try and use the series of $e^x$ here. $xe^x=x+\frac{x^2}{1!}+\frac{x^3}{2!}+\frac{x^4}{3!}...$

Differentiating with respect to $x$ 2 times we get: $xe^x+2e^x=\sum_{1}^{\infty}\frac{r(r+1)x^{r-1}}{r!}$

Finally, multiplying both sides by $x$ and dividing by 4 we get:

$\frac{xe^x(x+2)}{4}=\sum_{1}^{\infty}\frac{r(r+1)x^r}{4r!}$

All that remains to be done is to substitute $x=log9$ .

Upon doing so, the required expression becomes:

$9log3(1+log3)$ .

Answer= $\sqrt{9+3+3+1}=4$

It is easy to solve the problem once you know which series you need to manipulate and how do you need to do that. Begin with considering the well known expansion, $\displaystyle (1+x)^n = \sum_{r=0}^n \binom{n}{r} x^r$ . My motivation to choose this was that appearance of terms similar to that in binomial coefficients in the general term of the given series. Differentiating it w.r.t. $x$ , we obtain, $n(1+x)^{n-1} = \sum_{r=1}^n \binom{n}{r} r x^{r-1}$ To get the $r^2$ in the expression we need to do a similar task but with a slight modification. Multiply both the sides by $x$ again differentiate to get, $n\left((1+x)^{n-1} + (n-1)x(1+x)^{n-2} \right) = \sum_{r=1}^n \binom{n}{r} r^2 x^{r-1}$ Substitute $x=1$ to obtain, $\displaystyle \sum_{r=1}^n \frac{n!}{r!(n-r)!} r^2 = n(n+1) 2^{n-2} \Rightarrow a_n = \frac{n(n+1)}{n!} 2^{n-2} (\text{log } 3)^n$

So what we want is $\displaystyle S = \sum_{n=1}^{ \infty } a_n$ .

Again, the term $a_n$ looks very similar to a well known expression - the Maclaurin series for $e^x$ . Let us apply the procedure once again but with a slight modification. So, first multiply by $x$ and then differentiate. $\therefore \frac{\text{d }}{\text{d}x} \left( xe^x \right) = \frac{\text{d }}{\text{d}x} \left( \sum_{r=1}^{ \infty } \frac{x^{r+1}}{r!} + x \right)$ $\Rightarrow e^x(x+1) = \sum_{r=1}^{ \infty }(r+1) \frac{x^r}{r!} + 1$ Differentiate once again to obtain, $\Rightarrow e^x(x+2) = \sum_{r=1}^{ \infty }r(r+1) \frac{x^{r-1}}{r!}$ Substitute $x = 2 \text{ log }3$ to change the above expression into something that we want to have. $9\left(2 \text{ log }3 +2 \right) = \sum_{r=1}^{ \infty }r(r+1) \frac{\left( 2 \text{ log }3 \right)^{r-1}}{r!}$ $9\left(2 \text{ log }3 +2 \right) =\frac{2}{ \text{ log } 3} \sum_{r=1}^{ \infty } a_r$

$\displaystyle \therefore S = 18 (1+ \text{ log }3 ) \frac{\text{ log } 3}{2} \Rightarrow S = \left(1 + \text{ log } 3 \right)\left( 9 \text{ log } 3 \right)$ .

Compare with the given form to obtain the value of the constants and hence the answer $\displaystyle = \boxed{4}$ .