# Calculus Homework, Tutor couldn't help, I've hit the wall

• Nov 29th 2008, 07:20 AM
Harrigan
Calculus Homework, Tutor couldn't help, I've hit the wall
Solved
• Nov 29th 2008, 09:16 AM
galactus
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 $(k+1)^{7}-k^{7}=7k^{6}+21k^{5}+35k^{4}+35k^{3}+21k^{2}+7k+1$

$(n+1)^{7}-1=n^{7}+7n^{6}+21n^{5}+35n^{4}+35n^{3}+21n^{2}+7n$

Notice the coefficients?. They look binomial, don't they?

$\sum_{k=1}^{n}\left(7k^{6}+21k^{5}+35k^{4}+35k^{3} +21k^{2}+7k+1\right)$ $=n^{7}+7n^{6}+21n^{5}+35n^{4}+35n^{3}+21n^{2}+7n$

$7\sum_{k=1}^{n}k^{6}+21\sum_{k=1}^{n}k^{5}+35\sum_ {k=1}^{n}k^{4}+35\sum_{k=1}^{n}k^{3}+21\sum_{k=1}^ {n}k^{2}+7\sum_{k=1}^{n}k+\sum_{k=1}^{n}1$
$=n^{7}+7n^{6}+21n^{5}+35n^{4}+35n^{3}+21n^{2}+7n$

Now, plug in the respective identities for the powers. We must know them dong it this way.

$7\sum_{k=1}^{n}k^{6}+21\left(\frac{n^{2}(n+1)^{2}( 2n^{2}+2n-1)}{12}\right)+35\left(\frac{n(n+1)(6n^{3}+9n^{2}+ n-1)}{30}\right)$ $+35\left(\frac{n^{2}(n+1)^{2}}{4}\right)+21\left(\ frac{n(n+1)(2n+1)}{6}\right)+7\left(\frac{n(n+1)}{ 2}\right)+n$

$=n^{7}+7n^{6}+21n^{5}+35n^{4}+35n^{3}+21n^{2}+7n$

Now, solve for $\sum_{k=1}^{n}k^{6}$

It is lotsa horrible algebra, but if done correctly you should get:

$\sum_{k=1}^{n}k^{6}=\frac{1}{7}n^{7}+\frac{1}{2}n^ {6}+\frac{1}{2}n^{5}-\frac{1}{6}n^{3}+\frac{1}{42}n$

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.
• Nov 29th 2008, 10:05 AM
galactus

Quote:

The series of reciprocals of primes
Quote:

. 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.