Slippery Sum!

$S= \frac{1.2}{3}+ \frac{2.3}{3^2}+ \frac{3.4}{3^3}+ \dots$

Dividing both side by 3

$\frac{S}{3} = \frac{1.2}{3^2}+ \frac{2.3}{3^3}+ \frac{3.4}{3^4}+ \dots$

Subtracting second equation from the first,

$\frac{2S}{3} = \frac{1.2}{3}+ \frac{2.2}{3^2}+ \frac{3.2}{3^3}+\frac{4.2}{3^4}+ \dots$

$\Rightarrow \frac{S}{3}= \frac{1}{3}+ \frac{2}{3^2}+ \frac{3}{3^3}+\frac{4}{3^4}+ \dots \cdots(1)$

Again Dividing both side by 3 and subtracting from (1),

$\frac{2S}{9}=\frac{1}{3}+ \frac{1}{3^2}+ \frac{1}{3^3}+\frac{1}{3^4}+ \dots$

$\Rightarrow S= \frac{9}{2} \times \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{9}{4}$

So answer is $9+4=\fbox{13}$

Just to offer a different approach to Atul's solution, note first that

$\displaystyle\sum_{r=1}^{\infty} x^{r+1} = \dfrac{x^{2}}{1 - x}$ for $|x| \lt 1.$

Differentiating both sides with respect to $x$ , (the LHS term-by-term), and then simplifying, we have that

$\displaystyle\sum_{r=1}^{\infty} (r + 1)x^{r} = \dfrac{2x - x^{2}}{(1 - x)^{2}}.$

Differentiating again and simplifying, we find that

$\displaystyle\sum_{r=1}^{\infty} r(r + 1)x^{r} = \dfrac{2x}{(1 - x)^{3}}.$

Plugging in $x = \dfrac{1}{3}$ we find that $S = \dfrac{\frac{2}{3}}{(\frac{2}{3})^{3}} = \dfrac{9}{4},$

and thus $m + n = 9 + 4 = \boxed{13}.$

the expansion of ${ \left( 1-x \right) }^{ -n }$ is as follows

${ (1-x) }^{ -n }$ $=\quad 1\quad +\quad \left( \begin{matrix} n \\ 1 \end{matrix} \right) x\quad +\quad \left( \begin{matrix} n+1 \\ 2 \end{matrix} \right) { x }^{ 2 }\quad +\quad \left( \begin{matrix} n+2 \\ 3 \end{matrix} \right) { x }^{ 3 }\quad +......$ $=\quad 1\quad +\quad \left( \begin{matrix} n \\ n-1 \end{matrix} \right) x\quad +\left( \begin{matrix} n+1 \\ n-1 \end{matrix} \right) { x }^{ 2 }\quad +\quad \left( \begin{matrix} n+2 \\ n-1 \end{matrix} \right) \quad +....$

where $\left( \begin{matrix} n \\ r \end{matrix} \right)$ = $\frac { n! }{ (n-r)!r! }$ .

Putting n=3 we get

${ \left( 1-x \right) }^{ -3 }$ $=\quad 1\quad +\quad \left( \begin{matrix} 3 \\ 2 \end{matrix} \right) x\quad +\left( \begin{matrix} 4 \\ 2 \end{matrix} \right) { x }^{ 2 }\quad +\quad \left( \begin{matrix} 5 \\ 2 \end{matrix} \right) { x }^{ 3 }\quad +....$

Now multiplying both sides by 2x and substituting x=1/3 we get required sum.

${ 2\times 1/3\left( 1-1/3 \right) }^{ -3 }=\quad \frac { 1.2 }{ 3 } +\frac { 2.3 }{ { 3 }^{ 2 } } +\frac { 3.4 }{ { 3 }^{ 3 } } +...$

$=\frac { 9 }{ 4 }$

I would like to write more heuristic and mathematically correct solution

First of all, this sum looks like good candidate for power series in form

$S(x) = \displaystyle \sum_{n=0}^{+\infty} n (n+1) x^n \, ,$

for $x = \frac{1}{3}$ (there in no difference if we omit term for $n= 0$ or we don't). Here is important observation. Only power series with similar form we can directly calculate is geometric series

$f(x) = \displaystyle \sum_{n=0}^{+\infty} x^n = \frac{1}{1-x} \, ,$

which converges for $|x| < 1$ (this condition is fulfilled for $x = \frac{1}{3}$ ). Using derivative of power series, we could obtain the needed coefficient $n (n+1)$ .

For power series we know, that within radius of convergence of power series we can derive power series term by term.

When

$F(x) = \displaystyle \sum_{n=0}^{+\infty} a_n x^n \, ,$

for $x \in (-R,+R )$ , then

$\frac{\mathrm{d} F }{\mathrm{d} x } (x) = \displaystyle \sum_{n=0}^{+\infty} \frac{\mathrm{d} }{\mathrm{d} x } (a_n x^n) = \displaystyle \sum_{n=0}^{+\infty} a_n n x^{n-1} \, ,$

for $x \in (-R,+R )$ .

Now we do second derivative of $f(x)$ :

$f'(x) = \displaystyle \sum_{n=0}^{+\infty} n x^{n-1} = \displaystyle \sum_{n=1}^{+\infty} n x^{n-1} = \frac{1}{(1-x)^2} \, ,$

$f''(x) = \displaystyle \sum_{n=1}^{+\infty} n (n-1) x^{n-2} = \displaystyle \sum_{n=2}^{+\infty} n (n-1) x^{n-2} = \frac{2}{(1-x)^3} \, .$

Now we denote $m = n -1$ to obtain

$f''(x) = \displaystyle \sum_{m=1}^{+\infty} (m+1) m x^{m-1} = \frac{2}{(1-x)^3} \, .$

Especially for $x = \frac{1}{3}$

$\displaystyle \sum_{m=1}^{+\infty} (m+1) m \left( \frac{1}{3} \right)^{m-1} = \frac{27}{4} \, .$

Dividing equation with 3 we obtain answer

$\displaystyle \sum_{m=1}^{+\infty} \frac{m (m+1) }{3^{m-1}} \frac{1}{3} = \displaystyle \sum_{m=1}^{+\infty} \frac{m (m+1) }{3^m} = \frac{9}{4} \, ,$

and thus answer is $9 + 4 = 13$ .