Infinite series

Let $\begin{aligned}\begin{aligned}S=&\ 1+\dfrac45+\dfrac7{5^2}+\dfrac{10}{5^3}+\cdots\Rightarrow(1)\end{aligned}\end{aligned}$

Multiplying with $\dfrac15$ in equation $(1)$ ,

$\begin{aligned}\begin{aligned}\dfrac15S=&\ \dfrac15+\dfrac4{5^2}+\dfrac7{5^3}+\cdots\Rightarrow(2)\end{aligned}\end{aligned}$

Equation $(1)$ $-$ Equation $(2)$ ,

$\begin{aligned}\begin{aligned}\dfrac45S=&\ 1+\left(\dfrac35+\dfrac3{5^2}+\dfrac3{5^3}+\cdots\right)\\\dfrac45S=&\ 1+\left(\dfrac{\frac35}{1-\frac15}\right)\\\dfrac45S=&\ 1+\dfrac34\\\dfrac45S=&\ \dfrac74\\S=&\ \dfrac74\times\dfrac54=\boxed{\dfrac{35}{16}}\end{aligned}\end{aligned}$

I did it by simple logic.

The answer should be greater than the sum of first two terms , that is $1$ and $4/5$ , which is equal to $9/5$

We can observe that none of the fractions in the given options are greater than $9/5$ , except for $35/19$ .

Therefore the required answer is $35/19$ .

We see that the sequence is an arithmetico-geometric series. Let the sum be $S$ . We have:

$S=1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+\ldots$

$\frac{S}{5}=\frac{1}{5}+\frac{4}{5^2}+\frac{7}{5^3}+\ldots$

Subtracting the two, we get:

$\frac{4S}{5}=1+\frac{3}{5}+\frac{3}{5^2}+\frac{3}{5^3}+\ldots$

$\frac{4S}{5}=1+\cfrac{\frac{3}{5}}{1-\frac{1}{5}}=1+\frac{3}{4}=\frac{7}{4}$

We can multiply both sides by $\frac{5}{4}$ and get the desired sum of $S=\boxed{\frac{35}{16}}$

The infinite sum is $\displaystyle\sum_{n=1}^\infty \dfrac{3n-2}{5^{n-1}}=3\displaystyle\sum_{n=1}^\infty\dfrac{n}{5^{n-1}}-2\displaystyle\sum_{n=1}^\infty\dfrac{1}{5^{n-1}}$ Now let's differentiate our best friend :-) : $\displaystyle\sum_{n=0}^\infty x^n=\dfrac{1}{1-x}$ for $|x|<1$ so $\left(\displaystyle\sum_{n=0}^\infty x^{n}\right)'=\displaystyle\sum_{n=0}^\infty nx^{n-1}=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\dfrac{1}{(1-x)^2}$ For $x=\dfrac{1}{5}$ we have $\displaystyle\sum_{n=1}^\infty\dfrac{n}{5^{n-1}}=\dfrac{1}{\left(1-\dfrac{1}{5}\right)^2}=\dfrac{25}{16}$ Hence $\displaystyle\sum_{n=1}^\infty \dfrac{3n-2}{5^{n-1}}=3\displaystyle\sum_{n=1}^\infty\dfrac{n}{5^{n-1}}-2\displaystyle\sum_{n=1}^\infty\dfrac{1}{5^{n-1}}=3\cdot\dfrac{25}{16}-2\cdot\dfrac{1}{1-\dfrac{1}{5}}=\dfrac{35}{16}$ .