Solved
The sum of the 6th power identities, huh?. Well, here is my way. Years ago
when I figured this out, I thought I had done something. But, it would be silly
to think it was original, afterall.
Consider $\displaystyle (k+1)^{7}-k^{7}=7k^{6}+21k^{5}+35k^{4}+35k^{3}+21k^{2}+7k+1$
$\displaystyle (n+1)^{7}-1=n^{7}+7n^{6}+21n^{5}+35n^{4}+35n^{3}+21n^{2}+7n$
Notice the coefficients?. They look binomial, don't they?
$\displaystyle \sum_{k=1}^{n}\left(7k^{6}+21k^{5}+35k^{4}+35k^{3} +21k^{2}+7k+1\right)$$\displaystyle =n^{7}+7n^{6}+21n^{5}+35n^{4}+35n^{3}+21n^{2}+7n$
$\displaystyle 7\sum_{k=1}^{n}k^{6}+21\sum_{k=1}^{n}k^{5}+35\sum_ {k=1}^{n}k^{4}+35\sum_{k=1}^{n}k^{3}+21\sum_{k=1}^ {n}k^{2}+7\sum_{k=1}^{n}k+\sum_{k=1}^{n}1$
$\displaystyle =n^{7}+7n^{6}+21n^{5}+35n^{4}+35n^{3}+21n^{2}+7n$
Now, plug in the respective identities for the powers. We must know them dong it this way.
$\displaystyle 7\sum_{k=1}^{n}k^{6}+21\left(\frac{n^{2}(n+1)^{2}( 2n^{2}+2n-1)}{12}\right)+35\left(\frac{n(n+1)(6n^{3}+9n^{2}+ n-1)}{30}\right)$$\displaystyle +35\left(\frac{n^{2}(n+1)^{2}}{4}\right)+21\left(\ frac{n(n+1)(2n+1)}{6}\right)+7\left(\frac{n(n+1)}{ 2}\right)+n$
$\displaystyle =n^{7}+7n^{6}+21n^{5}+35n^{4}+35n^{3}+21n^{2}+7n$
Now, solve for $\displaystyle \sum_{k=1}^{n}k^{6}$
It is lotsa horrible algebra, but if done correctly you should get:
$\displaystyle \sum_{k=1}^{n}k^{6}=\frac{1}{7}n^{7}+\frac{1}{2}n^ {6}+\frac{1}{2}n^{5}-\frac{1}{6}n^{3}+\frac{1}{42}n$
I am sure there is a simpler way, but you can do any of them this way as long as you know the previous ones. Just plod through the algebra.
The series of reciprocals of primes. Here is a beautiful result proved by Euler in the year 1773. The result states that the series of all the reciprocals of primes diverges.
(1/2) + (1/3) + (1/5) + (1/7) + (1/11) + (1/13) ...
Your task is to prove this result. This is a bit of a tricky problem, so I will give you more elaborate hints. Assume to the contrary that the above series converges and arrive at a contradiction using the steps below. (Notation: p will denote a prime number)
This can probably be found on line somewhere. The outline of Euler's proof
of this can be found in "Euler, the Master of Us All" by William Dunham.