$E(X) = 2*(\frac{1}{2})^{1} + 3*(\frac{1}{2})^{2} + 4*(\frac{1}{2})^{3} + ......\infty$
It is the combination of AP and GP... Terms 2,3,4,... are in AP and Terms $\frac{1}{2} , (\frac{1}{2})^{2},...$ are in GP..
To solve these type of series , there is a fixed procedure...We multiply by the common ratio of GP (here it is 1/2) in both LHS and RHS..while multiplying by common ratio in RHS , we have to write by leaving space of one term and then subtract 2nd equation from 1st..
We repeat this procedure until we get the GP in RHS...
So, Here ,
$E(X) \;\;\;\;\;= \;\;\;\;\;\;2*(\frac{1}{2})^{1} + 3*(\frac{1}{2})^{2} + 4*(\frac{1}{2})^{3} + 5*(\frac{1}{2})^{4}+......\infty$
$\frac{1}{2}*E(X) = \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2*(\frac{1}{2})^{2} + 3*(\frac{1}{2})^{3} + 4*(\frac{1}{2})^{4} + ......\infty$
Now, vertically Subtract $2^{nd}$ equation from $1^{st}$ equation.
$(1-\frac{1}{2})*E(X)= 2*(\frac{1}{2})^{1}+ 1*(\frac{1}{2})^{2} + 1*(\frac{1}{2})^{3} + 1*(\frac{1}{2})^{4} + ......\infty$
$(\frac{1}{2})*E(X)= 2*(\frac{1}{2})^{1}+$ GP with first term as $(\frac{1}{2})^{2}$ and common ratio as 1/2...
$(\frac{1}{2})*E(X)$= 1 + $ \frac{(\frac{1}{2})^{2}}{(1-\frac{1}{2})}$