Sum of the first n terms of Central polygonal numbers (the Lazy Caterer's sequence)

Question

How do you find the sum of the first $n$ terms of the Central polygonal numbers (the Lazy Caterer's sequence): A000124 - OEIS

$1, 2, 4, 7, 11, 16, 22, 29, 37, 46, \ldots$

I tried but could not solve.

Please help!!!!

I am from Brazil.

Answer 1

There are many ways of deriving the $i$-th term.

One easy way is the following:

Notice that if the $i$-th term is $t_i$ then

$t_2 - t_1 = 1$

$t_3 - t_2 = 2$

$.$

$.$

$.$

$t_i - t_{i-1} = i-1$

Add all the above equations to obtain

$t_i - 1 = \frac{i(i-1)}{2}$

or, $t_i = \frac{1}{2}(i^2 - i + 2)$

Finally

$\displaystyle \sum_{i=1}^n t_i = \frac{1}{2} \left(\displaystyle \sum_{i=1}^n i^2 - \displaystyle \sum_{i=1}^n i + \displaystyle \sum_{i=1}^n 2 \right) = \frac{1}{2} \left[\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} + 2n\right] = \frac{n(n^2+5)}{6}$

Answer 2

You sequences seems to be defined as $\begin{cases}u_1=1\\u_{n+1}=u_n+n\end{cases}$

Thus $S(n+1)=\displaystyle\sum_{i=1}^{n+1} u_i=u_1+\sum_{i=2}^{n+1} u_i=1+\sum_{i=1}^{n}(u_i+i)=1+S(n)+\sum_{i=1}^{n}i=1+S(n)+\dfrac{n(n+1)}2$

This gives a telescopinc formula $S(n+1)-S(n)=1+\dfrac{n(n+1)}2$

Which sums to $$S(n+1)-S(1)=\sum_{i=1}^{n}\left(1+\dfrac{i(i+1)}{2}\right)=n+\dfrac{n(n+1)(n+2)}{6}$$

Since $S(1)=1$ we have $$S(n)=\dfrac{n(n^2+5)}{6}$$