2023/021) Find the sum of the series $1+5+10+16+23\cdots$

 Let us have a look at the $t^{th}$ term. 

Let us look at difference for a couple of terms

$t_2 = t_1 = 5- 1 = 4$

$t_3-t_2 = 10 -5 = 5$

$t_4-t_3 = 16 -10 = 6$

From the above we see that $t_k = t_{k-1} + k + 2$

From this let us find $t_n$  in terms of n

We have $t_n = 1 + \sum_{k=2}^{n} (k + 2)$

$= 1 + \frac{n(n+1)}{2} - 1 + 2(n-1)$

$= \frac{n^2}{2} + \frac{5n }{2} - 2$

So sum $S_n = \sum_{k=1}^n (\frac{k^2}{2} + \frac{5k }{2} - 2)$

$= \sum_{k=1}^n (\frac{k^2}{2}) + \sum_{k=1}^n\frac{5k }{2} - \sum_{k=1}^n2$

$=\frac{1}{2}(\frac{n(n+1)(2n+1)}{6}  + \frac{5}{2} \frac{n(n+1)}{2} - 2n $

$= \frac{2n^3 + 3n^2 +n }{12} + \frac{5n^2 +5n}{4} - 2n$

$= \frac{2n^3 + 3n^2 + n + 15n^2 + 15n -24n}{12}$

$= \frac{2n^3 + 18n^2 -8n}{12}$

$= \frac{n^3 + 9n^2 -4n}{6}$