Question

Please help me solve this​

Answer 1

You can factor the 32 out of the sum:

$\displaystyle \sum_{m=1}^\infty 32 \cdot \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1}$

We can also change the index as follows

$\displaystyle 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m}$

Now, we have a theorem that states that the series

$\displaystyle \sum_{m=1}^\infty a^m$

converges if and only if $|a|$, and in this case we have

$\displaystyle \sum_{m=1}^\infty a^m = \dfrac{1}{1-a}$

This is your case, because you have

$|a|=\dfrac{1}{2}$

which implies that your series converges, and the value is

$\displaystyle 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m} = 32 \cdot \dfrac{1}{1-\frac{1}{2}} = 32\cdot 2 = 64$