# Homework Help: Mathematical Induction

1. Jul 10, 2010

1. The problem statement, all variables and given/known data
Use mathematical induction to prove the formula for every positive integer n.

$$\sum_{i=1}^{5}i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}$$

2. Relevant equations

3. The attempt at a solution

I know this will be true because the RHS is just your standard sums of powers of integers to the fifth formula, but nevertheless I have to prove it...

I know n=1 will work so I set n=k

$$\sum_{k=1}^{5}k^5=\frac{k^2(k+1)^2(2k^2+2k-1)}{12}$$

I take the RHS and set k=(k+1)

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

If the test works for all the terms then it must be correct for the first 5 terms so I can eliminate the summation and just have the $$k^5$$

so then I'd have

$$(k+1)^5+\frac{k^2(k+1)^2(2k^2+2k-1)}{12}$$
$$=\frac{k^2(k+1)^2(2k^2+2k-1)+12(k+1)^5}{12}$$

From here I'm not sure how to get that to equal this:

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

2. Jul 10, 2010

### malicx

Your first formula as written is incorrect. The sum from 1 to 5 is not different for every positive integer, n. Your upper index should be n and not 5.

To use induction you must:
1. Verify the base case.
2. Assume that your case holds for some number, k.
3. Show that it holds for k+1. (IF it is true for k, THEN it is true for k+1)

Try adding $$(k+1)^5$$ to both sides of the equation and manipulate the right side until you get the proper form.

3. Jul 10, 2010

My first formula is written exactly as printed from my text book so I know it is correct. I know how to solve with induction my question is only one of algebra.

I need to get

$$\frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}=\frac{k^2(k+1)^2(2k^2+2k-1)+12(k+1)^5}{12}$$

The one on the left is where I subbed in k=(k+1) the one on the right is the equation I got from adding (k+1)5 to. I know they are equal because you can plug in a value and get the same number on both sides. How can you get these equations into the same form.

4. Jul 10, 2010

### LCKurtz

Failing anything clever, you could always expand and collect both numerators to see they are both:

2k6+18k5+65k4+120k3+119k2+60k+12

5. Jul 10, 2010

### ehild

That is correct. Note that both sides have the common factor (k+1)^2, so you have to prove the simpler equality

$$\frac{(k+2)^2(2(k+1)^2+2k+1)}{12}=\frac{k^2 (2k^2+2k-1)+12(k+1)^3}{12}$$.

After that you can follow LCKurtz suggestion to expand both sides and see if they are identical.

ehild

6. Jul 11, 2010

Alright, that works.

Thanks.

7. Jul 11, 2010

### malicx

At the risk of sounding like a stickler/jerk, the formula is NOT correct, even if it is what is listed in the book. Typically, a formula in n will be given when summing from 1 to n, but that is summing from 1 to 5.

8. Jul 11, 2010

### ehild

You are right, it must be a typo. So the correct form is

$$\sum_{i=1}^{n}i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}$$

ehild