Proving the Formula Using Mathematical Induction

[themadhatter1]

Homework Statement

Use mathematical induction to prove the formula for every positive integer n.

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

Homework Equations

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:

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

 

[malicx]

Homework Statement

Use mathematical induction to prove the formula for every positive integer n.

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

Homework Equations

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:

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

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.

 

[themadhatter1]

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.

My first formula is written exactly as printed from my textbook 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.

 

[LCKurtz]

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

2k⁶+18k⁵+65k⁴+120k³+119k²+60k+12

 

[ehild]

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

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

 

[themadhatter1]

Alright, that works.

Thanks.

 

[malicx]

My first formula is written exactly as printed from my textbook so I know it is correct.

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.

 

[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