Can Induction Prove the Sum of i^5 from 1 to n?

[Tom McCurdy]

I was trying to help a friend do the following problem

Prove with induction

Sum of $$i^5$$ from 1 to n =

$$\frac{n^2(n+1)^2(2n^2+2n-1}{12}$$

we got it to

$$\frac{(k+1)^2(k+2)^2(2k^2+4k+2)+2k+1}{12}$$ but we can't seem to get it to go back to the orginal equation when you substitue k+1 into the orignal formula

 

[vincentchan]

you might made an algebra mistake

try plug an integer into k... you'll see the numerator is odd and the denominator is even, which mean the answer is not even an integer...

check your algebra....

 

[wais]

First, it is important to note that the original formula may have a typo and should be \frac{n^2(n+1)^2(2n^2+2n+1)}{12} instead of \frac{n^2(n+1)^2(2n^2+2n-1}{12}. I will proceed with the assumption that this is the correct formula.

To prove this formula using induction, we need to show that it holds true for the base case (n=1) and then prove the inductive step, which is to show that if it holds true for n=k, then it also holds true for n=k+1.

Base case (n=1):
When n=1, the formula becomes \frac{1^2(1+1)^2(2(1)^2+2(1)+1)}{12} = \frac{1(4)(3)}{12} = \frac{12}{12} = 1. This is indeed the sum of i^5 from 1 to 1, so the formula holds true for the base case.

Inductive step:
Assuming the formula holds true for n=k, we need to show that it also holds true for n=k+1. Substituting k+1 into the original formula, we get:

\frac{(k+1)^2((k+1)+1)^2(2(k+1)^2+2(k+1)+1)}{12} = \frac{(k+1)^2(k+2)^2(2k^2+4k+3)}{12}

Expanding this expression, we get:

\frac{(k^2+2k+1)(k^2+4k+4)(2k^2+4k+3)}{12} = \frac{2k^6+12k^5+30k^4+44k^3+38k^2+17k+3}{12}

Using the formula given in the problem, we can rewrite this expression as:

\frac{2k^6+12k^5+30k^4+44k^3+38k^2+17k+3}{12} = \frac{(k+1)^2(k+2)^2(2k^2+4k+2)+2k+1}{12}

We can see that