The infinite sum problem How to calculate the following sum

$$1 + (4/7) + (9/7^{2}) + (16/7^{3}) + (25/7^{4}) + \ldots$$

I dont have prior experience to deal with kinds of problem.
$${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\quad {\text{ for }}|x|&lt;1\!$$
differentiate  it
$${\frac {1}{(1-x)^2}}=\sum _{n=0}^{\infty }nx^{n-1}$$
multiply by $x$
$${\frac {x}{(1-x)^2}}=\sum _{n=1}^{\infty }nx^{n}$$
differentiate it
$$({\frac {x}{(1-x)^2}})'=\sum _{n=1}^{\infty }n^2x^{n-1}$$

$$\frac{1+x}{(1-x)^3}=\sum _{n=1}^{\infty }n^2x^{n-1}$$
at $x=\frac{1}{7}$
$$\frac{1+\frac{1}{7}}{(1-\frac{1}{7})^3}=\frac{49}{27}$$
Hint: For $|x|&lt;1$ $$\sum_{k=0}^{\infty}x^k=\frac{1}{1-x}$$Differentiate to get: $$ \sum_{k=0}^{\infty}kx^k=\frac{x}{(1-x)^2}$$ Differentiate once again to obtain $\sum_{k=0}^{\infty}k^2x^k$: $$\sum_{k=0}^{\infty}k^2x^k=\frac{x(1-x+x^2)}{(1-x)^3} $$ Here $|x|=1/7&lt;1$, so we can use the above result.