$1, 2,2,3,3,3,4,4,4,4,5,5,5,5,5,\ldots ,n$ $(n$ times$)$
Approach:
Observation: For summation, let us consider a term $t_x$ as the sum of the terms which follow a general order.
Like:
$$t_1 = 1+2+3+4+5+6+\ldots+n$$
$$t_2 = 2+3+4+5+6+\ldots+n$$
$$t_3 = 3+4+5+6+\ldots+n$$
$$.$$
$$.$$
$$.$$
$$t_{n-1} = n -1 + n$$
$$t_n = n$$
For complete summation, we need to add all the $t_x$ obtained from the above.
For this particular step, I am getting a really big and lengthy sigma, which I am not able to evaluate.
$$\sum_{n=1}^n t_n$$
Now each term $t_n$ comprises of a summation of its very own.
$$\sum t_n = \sum_{n=k}^n \cfrac{n(n+1)}{2}$$
where $k= 1,2,3,\ldots,n$.
What to do next? After plugging in values for $k$, I am getting a really ugly expression. Please help.
P.S: This question might have been asked before, but I want to solve it through the above mentioned method. So please don’t mark it as duplicate and close it.
Best Answer
$$\sum\limits_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$