Q.Compute the sum of first $n$ terms of $1+(1+4)+(1+4+4^2)+(1+4+4^2+4^3)+\cdots$.

A finite series is the sum of the terms of a finite sequence: if $(a_n)$ is a sequence, $a_1+a_2+\cdots+a_n=\sum_{k=1}^n a_k$ is the corresponding finite series (sometimes indexed to start from $a_0$).

Sum of an AP. For $T_n=a+(n-1)d$,

Sum of a GP. For $T_n=ar^{n-1}$ ($r\ne1$),

if $r=1$ the GP is the constant sequence $a,a,\ldots$ and $S_n=na$. Equivalently $1+r+r^2+\cdots+r^{n-1}=\dfrac{1-r^n}{1-r}$ for $r\ne1$.

Sum of an AGP. For $T_n=(a+(n-1)d)r^{n-1}$ ($r\ne1$),

This is derived the same way as the GP-sum formula: write $S_n$ and $rS_n$, subtract, and simplify the resulting mixed AP/GP remainder.

Telescopic summation. Many series that are not AP, GP or AGP can still be summed exactly if the $k^{th}$ term can be rewritten as a difference $t_k=f(k)-f(k+1)$ (often via rationalising a surd denominator, or a partial-fraction split). Then

$\sum_{k=1}^n t_k = \big(f(1)-f(2)\big)+\big(f(2)-f(3)\big)+\cdots+\big(f(n)-f(n+1)\big) = f(1)-f(n+1),$ …