how do you actually prove this?

>>8535862
Induction.

>>8535868
but where do you start your induction? do you associate the product of two sums of squares and just multiple them together?

File: Untitled.png (56KB, 1400x702px) Image search: [Google]

56KB, 1400x702px

also quintics are terrifying. i don't know nearly enough math to touch this topic.

>>8535862
You prove that sum[i=1..n]f(i)=g(n) by showing that the difference g(n)-g(n-1)=f(n).

sum[i=1..n]f(i) = g(n)
=> sum[i=1..n-1]f(i) = g(n-1)
=> g(n)-g(n-1) = sum[i=1..n]f(i) - sum[i=1..n-1]f(i)
= f(n)

If you need to /find/ the expression for the sum of a polynomial, you start with the fact that (n+1)^k-n^k =
(n^k+k.n^(k-1)+...)-n^k = k.n^(k-1)+..., i.e. the highest-degree terms will always cancel so the sum will be a polynomial of degree one higher than the individual terms.

Then it's just a matter of expanding out g(n)-g(n-1)=f(n) where g(n) is the generic degree-k+1 polynomial, equating coefficients, and solving the resulting system of linear equations.

>>8535862
this is trivial by just using the same step 5 times, then simplify:
induction, let n= n+1:
n(1+n)(1+2n)(-1+3n+3n^2) -> (n+1)(1+n+1)(1+2(n+1))(-1+3(n+1)+3(n+1)^2)
=n(1+n+1)(1+2(n+1)(-1+3(n+1)+3(n+1)^2)) + (1+n+1)(1+2(n+1)(-1+3(n+1)+3(n+1)^2))
=n(1+n)(1+2(n+1)(-1+3(n+1)+3(n+1)^2)) + n(1+2(n+1)(-1+3(n+1)+3(n+1)^2)) + (1+n+1)(1+2(n+1)(-1+3(n+1)+3(n+1)^2))
=n(1+n)(1+2n)(-1+3(n+1)+3(n+1)^2)) + n(1+n)(2)(-1+3(n+1)+3(n+1)^2))+ n(1+2(n+1)(-1+3(n+1)+3(n+1)^2)) + (1+n+1)(1+2(n+1)(-1+3(n+1)+3(n+1)^2))
=n(1+n)(1+2n)(-1+3n+3(n+1)^2)) + n(1+n)(2)(3)+ n(1+2(n+1)(-1+3(n+1)+3(n+1)^2)) + (1+n+1)(1+2(n+1)(-1+3(n+1)+3(n+1)^2))
=n(1+n)(1+2n)(-1+3n+3n^2)) +n(1+n)(1+2n)(-1+3n+3(2n+1))) + n(1+n)(2)(3)+ n(1+2(n+1)(-1+3(n+1)+3(n+1)^2)) + (1+n+1)(1+2(n+1)(-1+3(n+1)+3(n+1)^2))

then simplify
n(1+n)(1+2n)(-1+3n+3(2n+1))) + n(1+n)(2)(3)+ n(1+2(n+1)(-1+3(n+1)+3(n+1)^2)) + (1+n+1)(1+2(n+1)(-1+3(n+1)+3(n+1)^2)) = 30(n+1)^4

QED

probably made some mistake in there somewhere wit the brackets, but the steps are correct.

look for a 5th degree polynomial P(x) such that
P(0)=0
P(1)=1
P(2) = 1+2^4
...
P(5) = 1+2^4 + 3^4 +4^4+5^4
. this uniquely defines the polynomial.
now check by induction

>>8536069
ehhh bit sloppy saying let n=n+1 and not including a base case.

but essentially this is right. just show that it holds when n=1 and when it holds for some n then it will hold for n+1

i.e. 1/30n(1+n)(1+2n)(3n^2+3n-1)+(n+1)^4=1/30(n+1)(1+n+1)(1+2(n+1))(3(n+1)^2+3(n+1)-1)

if you didnt know the equation in the first place then you'd give it the general solution an^5+bn^4+cn^3+dn^2+en (+f but f=0 by plugging in n=0). get 5 equations by plugging in values and then use gaussian elimination.

More elegantly, you can use discrete calculus. This is how the explicit formula for any sum of powers is derived.

Find a polynomial [math]P[/math] of degree 5 such that [math]P \left( X \,+\, 1 \right) \,-\, P \left( X \right) \,=\, X^4[/math] (hint: the constant coefficient doesn't matter). Once you find it, check if [math]P \left( X \,+\, 1 \right) \,-\, P \left( X \right) \,=\, X^4[/math] (otherwise you'll have to check the final result by induction which is the most retarded idea ever).
Sum [math]\sum_{n \,=\, 1}^n \left[ P \left(k \,+\, 1 \right) \,-\, P \left( k \right) \right][/math] as a telescopic sum.
Factorize your shit.

>>8535862
Define
[math] P_j(n)=\sum_{k=1}^n k^j [/math]

and
[math] S(t)=\sum_{j=0}^{\infty} t^jP_j(n)/j![/math]

Then
[math]S(t)=\sum_{k=1}^n\sum_{j=0}^{\infty} (tk)^j/j![/math]

[math]\sum_{k=1}^n e^{tk}={\frac {1-e^{tn}}{e^{-t}-1}}[/math]

To find P_4(n), simply extract the coefficient of t^4 in the corresponding Taylor series

In other words

[math] P_4(n)=({\frac {d^4} {dt^4}}\mid_{t=0}){\frac {1-e^{tn}}{e^{-t}-1}} [/math]

you can verify that this gives you the formula in the pic using wolfram alpha

>>8535862
https://en.wikipedia.org/wiki/Faulhaber%27s_formula

>>8537659
Holy shit, I love this one.
Here is one more:
Expand
[math] (k+1)^5-k^5[/math]
and sum from k=1 to k=n. As long as you know the formula for exponents k=1,2,3 (which you find with similar method) you're set.

>>8537978
And here's another:
Look at this formula:
[math]\sum_k=4^n\binom{n}{4}=\binom{n+1}{5}[/math]
Once again once you know the formula for the exponents 1,2,3 (found with similar method) you're set.
The formula above can be proven with combinatorics:
Right hand side says choose 5 people from n+1.
On the left you count it in other way, assume you have them in list (say alphabetic). First decide the last one ([math]k=5,\dots,n+1[/math])
and choose 4 from the ones before.

>>8537988
Fuck.
Screwed the formula
[math]\sum_{k=4}^n\binom{k}{4}=\binom{n+1}{5}[/math]

>>8535874

It kinda varies from sum to sum. Fortunately, sums over a term like k^i where i is some integer are easy to deduce. Summing them is a lot like taking the integral, so the integral of k^n is over order k^(n+1). So, we can just write the general n+1 polynomial, and then generate enough terms to determine all the coefficients.

Then you use induction.