Serious stuff for July.
1. Sum of n even squares.
This is related to the sum of the squares of the integers. Lots of different proofs; how about we start from the difference of consecutive cubes.
[tex]\displaystyle S = \sum_{k=1}^n ((k+1)^3 - k^3) [/tex]
[tex]\displaystyle S = \sum_{k=1}^n (3k^2 + 3k + 1) [/tex]
[tex]\displaystyle S = 3 \sum k^2 + 3 \sum k + \sum 1[/tex]
[tex]\displaystyle S = 3 \sum k^2 + 3n(n+1)/2 + n[/tex]
[tex]\displaystyle \sum k^2 = \frac 1 3(S - (3/2) n(n+1) - n)[/tex]
So all we need is S which is nicely telescoping:
[tex] S = (2^3 - 1^3) + (3^2 - 2^3) + (4^3 - 3^3) + ... + (n+1)^3-n^3 [/tex]
[tex] S = (n+1)^3 - 1 [/tex]
[tex]\displaystyle \sum k^2 = \frac 1 3((n+1)^3 - 1 - (3/2) n(n+1) - n)[/tex]
[tex]\displaystyle \sum k^2 = \frac 1 6 n (n + 1) (2 n + 1)
[/tex]
We're interested in
[tex]\displaystyle \sum (2k)^2 = 4 \sum k^2 = \frac 2 3 n (n + 1) (2 n + 1)
\quad\checkmark[/tex]
That's just the first one. This is gonna be a long problem set.
2.
[tex]\displaystyle S = \sum_{j=1}^n (3j - 2)[/tex]
That's a lot easier.
[tex]\displaystyle S = 3 \sum j - \sum 2[/tex]
[tex]S = 3n(n+1)/2 - 2n[/tex]
[tex]S = 3n(n+1)/2 - 2n[/tex]
[tex]S = \dfrac 1 2 (3n^2 -n) \quad\checkmark[/tex]
3.
[tex]\displaystyle \sum_{k=1}^n (5k-1) = 5 \sum k - \sum 1 = \frac 5 2 n(n+1)-n = \frac n 2(5n+3) \quad\checkmark [/tex]
4.
Now it's getting interesting again.
[tex]\displaystyle S=\sum_{p=1}^n p(2^{p-1}) [/tex]
If I recall, the trick with these is turning them into double sums. Instead of multiplying by p, we'll add up p times.
[tex]\displaystyle S=\sum_{p=1}^n \sum_{q=1}^p 2^{p-1} [/tex]
Let's just look at this for a minute as
[tex]\displaystyle S=\sum_{p=1}^n \sum_{q=1}^p f(p) = f(1) + f(2)+f(2) + f(3)+f(3)+f(3)+...[/tex]
This is the tricky part; we want to exchange the order of the sums but we have the outer index p in the inner bound. When q is the outer sum it can't go to p so it must go to n:
[tex]\displaystyle S=\sum_{q=1}^n \sum_{p=?}^? f(p) [/tex]
We have to fill in the question marks. We need [tex]1f(2), 2f(2), ...nf(n)[/tex] so we can see the bounds are:
[tex]\displaystyle S=\sum_{q=1}^n \sum_{p=q}^n f(p) [/tex]
We can write the sum as the difference of two sums from 1.
[tex]\displaystyle S= \sum_{q=1}^n \left(\sum_{p=1}^n f(p) - \sum_{p=1}^{q-1} f(p) \right) [/tex]
OK we substitute back in for f.
[tex]\displaystyle S=\sum_{q=1}^n \left( \sum_{p=1}^n 2^{p-1} - \sum_{p=1}^{q-1} 2^{p-1} \right) [/tex]
That's two geometric series
[tex]\displaystyle S=\sum_{q=1}^n ( (1-2^n)/(1-2) - (1-2^{q-1})/(1-2) ) [/tex]
[tex]\displaystyle S=\sum_{q=1}^n ( 2^n - 2^{q-1}) [/tex]
[tex]\displaystyle S=\sum_{q=1}^n 2^n - \sum_{q=1}^n 2^{q-1}[/tex]
[tex]\displaystyle S=n 2 ^n - (1-2^n)/(1-2)[/tex]
[tex]\displaystyle S = (n-1)2^n + 1 \quad\checkmark[/tex]
5.
[tex]\displaystyle S = \sum_{k=1}^n \dfrac{1}{k^2 + k} [/tex]
Hmmm. Looks complicated but is that just a telescoper?
[tex]\displaystyle S = \sum_{k=1}^n \dfrac{1}{k(k+1)} [/tex]
It's not too hard to just guess the partial fractions:
[tex]\displaystyle S = \sum_{k=1}^n \left(\dfrac{1}{k} - \dfrac{1}{k+1}\right) [/tex]
[tex]\displaystyle S = \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... + \dfrac{1}{n} - \dfrac{1}{n+1} [/tex]
[tex]\displaystyle S = 1 - \dfrac{1}{n+1} [/tex]
[tex]\displaystyle S = \dfrac{n}{n+1} \quad\checkmark [/tex]
Phew. Tough problem set. It occurs to me now we could have done the hard ones by induction which probably would have been simpler.
