Series:
5, 11, 19, 29, 41
Find the sum of the series up to $n$ terms.
Well the method that comes to my mind is to find the nth term of the sequence, and then find their summation. I use the basic formulas, such as sum of series $n^2$, $n^3$, etc.
My question is whether their is a shorter method to achieving this, or maybe just an alternate method.
Best Answer
Let $\displaystyle S=5+11+19+29+41+\cdots+T_{n-1}+T_n$
$\displaystyle S=5+11+19+29+41+\cdots+T_{n-1}+T_n$
$\displaystyle S-S=5+\underbrace{(11-5)+(19-11)+(29-19)+(41-29)+(T_n-T_{n-1})}-T_n$
$\displaystyle\implies T_n-5=6+8+10+12$ upto $n-1$terms $=(n-1)\{2\cdot6+(n-2)2\}=(n-1)(n+4)$
$\displaystyle T_n=5+n^2+3n-4=n^2+3n+1$
$\displaystyle\implies S_n=\sum_{r=1}^nr^2+3\sum_{r=1}^nr+\sum_{r=1}^n1$