Sequence and Series 1

$\displaystyle \sum^{\infty}_{n=1}\dfrac{n}{2}\left( \dfrac{1}{2^n} + \dfrac{1}{3^n}\right)=\frac{1}{2}\sum^{\infty}_{n=1}( \dfrac{n}{2^n} + \dfrac{n}{3^n})=\frac{1}{2}[\sum^{\infty}_{n=1}( \dfrac{n}{2^n}) +(\sum^{\infty}_{n=1}\dfrac{n}{3^n})]$ $\sum^{\infty}_{n=1}( \dfrac{n}{2^n})=S_1=( \dfrac{1}{2^1})+( \dfrac{2}{2^2})+( \dfrac{3}{2^3})+( \dfrac{4}{2^4})...~~(1)$ $~~~~~~~~~~~~~~~~~~~~\frac{S_1}{2}=( \dfrac{1}{2^2})+( \dfrac{2}{2^3})+( \dfrac{3}{2^4})+( \dfrac{4}{2^5})...~~(2)$ ~ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\frac{S_1}{2}=( \dfrac{1}{2^1})+( \dfrac{1}{2^2})+( \dfrac{1}{2^3})+( \dfrac{1}{2^4})...~~(1)-(2)~~\color{#D61F06}{{GP}}$ ~ ${\boxed{S_1=2}}$ $\sum^{\infty}_{n=1}(\frac{n}{3^n})=S_2=(\frac{1}{3^1})+(\frac{2}{3^2})+(\frac{3}{3^3})+(\frac{4}{3^4})...~~(3)$ $~~~~~~~~~~~~~~~~~~~\frac{S_2}{3}=(\frac{1}{3^2})+(\frac{2}{3^3})+(\frac{3}{3^4})+(\frac{4}{3^5})...~~(4)$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\frac{2S_2}{3}=(\frac{1}{3^1})+(\frac{1}{3^2})+(\frac{1}{3^3})+(\frac{1}{3^4})...~~~(3)-(4)~~\color{#20A900}{{GP}}$ ${\boxed{S_2=\frac{3}{4}}}$ $\frac{1}{2}[\sum^{\infty}_{n=1}( \dfrac{n}{2^n}) +(\sum^{\infty}_{n=1}\dfrac{n}{3^n})]=\frac{1}{2}[S_1+S_2]=\frac{1}{2}\times[\frac{11}{4}]=\frac{11}{8}=\frac{A}{B}$ $\large\color{#3D99F6}{\boxed{A+B=19}}$

Desired terms can be separated $\small{\color{#624F41}{\text{(To form AGP's)}}}$ as $\color{#D61F06}{\dfrac{1}{2}( (1)\dfrac{1}{2} +( 2)( \dfrac{1}{2^2})+(3 )( \dfrac{1}{2^3})+ }\ldots$ $\color{forestgreen}{+\dfrac{1}{2}( (1)\dfrac{1}{3} +( 2)( \dfrac{1}{3^2})+(3 )( \dfrac{1}{3^3})+} \ldots$ For $\color{#D61F06}{red}$ series (S) $S=\dfrac{1}{2}( (1)\dfrac{1}{2} +( 2)( \dfrac{1}{2^2})+(3 )( \dfrac{1}{2^3})+ \ldots$ $\dfrac{S}{2}=\dfrac{1}{2}( (1)\dfrac{1}{2^2} +( 2)( \dfrac{1}{2^3})+(3 )( \dfrac{1}{2^4})+ \ldots$ Subtracting, we get $\dfrac{S}{2}=\dfrac{1}{2}(\dfrac{1}{2} +\dfrac{1}{2^2}+\dfrac{1}{2^3}+ \ldots$ \Rightarrow S=1\quad\quad\small{\color{}{\text{(Sum of a Infinite GP)}}} In a similar fashion we can obtain sum of $\color{forestgreen}{green}$ AGP and comes out to be $\frac{3}{8}$ . Hence required sum= $\color{#D61F06}{1}+\color{forestgreen}{(\frac{3}{8})}=\frac{11}{8}$ .

Hence, $\large 8+11=19$