So at my school today, we celebrated Pi Day, and we had a customary mini-olympiad. In that olympiad there was a really interesting question.
There is a sequence: $\lbrace 1\rbrace, \lbrace 4, 9\rbrace, \lbrace 16, 25, 36 \rbrace, \lbrace 49, 64, 81, 100 \rbrace ...$ Find a closed form for $S$ where $S$ is the sum of the numbers in the nth set.
So a few observations that I made:
- The numbers are all perfect squares (pretty obvious, I know)
- The root of the last term of the nth set is $n$ more than the root of the last term of the previous set.
Well, I think that Gerry Myerson is right. Here are some of the instruction.
First is to determine the starting number of set nth (number without square). Let call this as $u_n$.
$u_1 = 1, u_2 = 2, u_3 = 4...$
Let assume $u_n = \frac{n(n-1)}{2}+1$
Noticing that set nth will have $n$ elements, then:
$u_{n+1} = (u_n + n-1) + 1 = \frac{n(n-1)}{2}+1 + n = \frac{n(n+1)}{2}+1$
Therefore, the starting number in the nth set is $[\frac{n(n-1)}{2}+1]^2$. Then:
$S_n = \sum_{k=\frac{n(n-1)}{2}+1}^{\frac{n(n+1)}{2}}k^2 = \sum_{k=1}^{\frac{n(n+1)}{2}} k^2- \sum_{k=1}^{\frac{n(n-1)}{2}}k^2$
And the sum of nth square number is given as:
$\sum_{k=1}^{k=n}k^2 = \frac{n(n+1)(2n+1)}{6}$
Now, you can substitute and simplify to get the result.