squares and the sevens

Let $~~x=1+\dfrac{4}{7}+\dfrac{9}{7^2}+\dfrac{16}{7^3}+\ldots\quad (1)\rightarrow \text{multiply (1) by}~\frac{1}{7} \\ \dfrac{1}{7}x=~~~~~\dfrac{1}{7}+\dfrac{4}{7^2}+\dfrac{9}{7^3}+\ldots\quad ~(2)$

Subtracting $(1)~ \text{to}~(2)$ , we have

$\dfrac{6}{7}x=1+\dfrac{3}{7}+\dfrac{5}{7^2}+\dfrac{7}{7^3}+\ldots\quad ~(3)\rightarrow \text{again, multiply (3) by}~\frac{1}{7}\\ \dfrac{6}{49}x=~~~~~\dfrac{1}{7}+\dfrac{3}{7^2}+\dfrac{5}{7^3}+\ldots\quad ~(4)$

again, subtracting $(3)~ \text{to}~(4)$ , so

$\dfrac{36}{49}x=1+\left(\dfrac{2}{7}+\dfrac{2}{7^2}+\dfrac{2}{7^3}+\ldots\right)$

In the last equation, on RHS in bracket, we have an infinite geometric progression with first term $\frac{2}{7}$ and common ratio $\frac{1}{7}$ . Thus

$\dfrac{36}{49}x=1+\dfrac{\frac{2}{7}}{1-\frac{1}{7}}=\dfrac{4}{3}\\x=\dfrac{4\times{49}}{3\times{36}}=\boxed{\dfrac{49}{27}}$

But, otherhand, we can use Arithmetic-Geometric Progression on the equation $(3)$ . Based on Arithmetic-Geometric Progression on RHS in equation $(3)$ , we have $a=1, r=\dfrac{1}{7}, \text{and}~d=2$ . Then,

$\dfrac{6}{7}x=\dfrac{1}{1-\frac{1}{7}}+\dfrac{{2}\times{\frac{1}{7}}}{\left(1-\frac{1}{7}\right)^2}=\dfrac{7}{6}+\dfrac{14}{36}=\dfrac{56}{36}\\x=\dfrac{56\times{7}}{36\times{6}}=\boxed{\dfrac{49}{27}}$

Solution by Mas Mus is no doubt very cool. But to think of such a solution is not so easy. I use TI-83 Plus calculator and set as follows. $1 ~STR~A~ENTER.~1~ STR~X~ENTER.\\X+1~ STR~ X~ : A+\dfrac{7*X^2}{7^n}~ STR~ A$
press ENTER m times and we get sum of (m+1) terms. Aftter m=15, I got the sum as 1.814814815 for next several pressings of ENTER.. Now this suggested .814814815 is actually recurring 0.814. And that is $\dfrac {22}{27}. ~So~ the~ answer~ is~ 1\dfrac{22}{27}=\dfrac {49}{27}.$
Logically, we see that after m=15, the term added is in the order of $E^{-10}$ . So at m=15 we almost get the sum. Beyond it is out of calculator accuracy.

Here is a solution that use power series, of course it's quite a big tool for this but this is an alternative to the (better) solution already given :)

Let us name $S$ the result, and define on $]-1,1[$ f the following power series $f(x) = \sum_{n=0}^\infty n^2x^n$ such that we have $S=f(1/7)$ .
Let us now define on $]-1,1[$ g such as $g(x) = \sum_{n=0}^\infty x^n$ One can easily check that $g(x) = \frac{1}{1-x}$ (by seeing that $1+x+x^2+...+x^N)(1-x) = 1-x^N$ , and making $N$ go to infinity) Also, $f'(x) = \frac{1}{(1-x)^2} = \sum_0^\infty nx^{n-1} = \sum_1^\infty nx^{n-1}$ $= \sum_1^\infty(n-1)x^{n-1}+\sum_1^\infty x^{n-1} = \sum_0^\infty nx^{n}+f(x)$ Then $f''(x) = \frac{2}{(1-x)^3} = g(x) + f'(x)$ , and we just need to evaluate $g$ in $1/7$ , as we now have the following expression :

$g(x) = \frac{2}{(1-x)^3} - \frac{1}{(1-x)^2} = \left(\frac{2}{1-x}-1\right)\frac{1}{(1-x)^2}$

$g(1/7) = (2 \cdot 7/6 - 1 ) (7/6)^2= \frac{8 \cdot 49}{6 \cdot 36} = \frac{2^3\cdot 7^2}{2^3\cdot3^3} = \frac{49}{27}$