1. Discrete sigma

Hi

Consider the following similarities between the sums:

1 = 1
2 + 3 + 4 = 1 + 8
5 + 6 + 7 + 8 + 9 = 8 + 27
10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64

Do you see the pattern? Find a general formula, with sigma and Exponentiations, which summarizes this pattern.

I need some guiding...

2. Originally Posted by iHeji
Hi

Consider the following similarities between the sums:

1 = 1
2 + 3 + 4 = 1 + 8
5 + 6 + 7 + 8 + 9 = 8 + 27
10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64

Do you see the pattern? Find a general formula, with sigma and Exponentiations, which summarizes this pattern.

I need some guiding...
Perhaps this will help, perhaps it will not.

The first equation (N = 1) is given as
$\displaystyle \sum_{n=1}^{1+0}n = (N - 1)^3 + N^3$

The second (N = 2) is
$\displaystyle \sum_{n=2}^{2+2}n = (N - 1)^3 + N^3$

The third (N = 3) is
$\displaystyle \sum_{n=5}^{5+4}n = (N - 1)^3 + N^3$

etc.

Notice that the lowest number for n is given by the series 1, 2, 5, 10, 17, etc. This is the recursive function
$f(N) = f(N - 1) + (2N - 3),~f(1) = 1$

where N is the equation number. So we have:
The first equation (N = 1) is given as
$\displaystyle \sum_{n=f(1)}^{f(1)+0}n = (N - 1)^3 + N^3$

The second (N = 2) is
$\displaystyle \sum_{n=f(2)}^{f(2)+2}n = (N - 1)^3 + N^3$

The third (N = 3) is
$\displaystyle \sum_{n=f(3)}^{f(3)+4}n = (N - 1)^3 + N^3$

etc.

Now notice that the upper term of the summation is of the form f(N) + g(N) where g(n) = 2(N - 1). So the general Nth equation is given by

$\displaystyle \sum_{n = f(N)}^{f(N) + 2(N - 1)}n = (N - 1)^3 + N^3$

This is probably a more complicated method than is needed, but it's the one I came up with that was the most demonstrative. Anyway, now solve f(N) as a function of N (rather than by recursion) and I'd guess the rest would be a simple induction proof.

-Dan

3. Go figure. That F(N) function is a simple quadratic. I should have seen that coming.

-Dan

4. I just finished this. It's takes a while to get an explicit form for f(N), but it is do-able this way. Perhaps one of the other helpers will have a shortcut.

-Dan