# Thread: Prove that the sum of fourth powers of the first n integers is 1/30n(n+1) (2n+1)(

1. ## Prove that the sum of fourth powers of the first n integers is 1/30n(n+1) (2n+1)(

Prove that the sum of fourth powers of the first n integers is 1/30n(n+1) (2n+1)(3n2+3n+1)

2. Originally Posted by rajeev000
Prove that the sum of fourth powers of the first n integers is 1/30n(n+1) (2n+1)(3n2+3n+1)
sort of confused about the formula you wrote, since i am sure there are brackets missing that would be needed to clarify. nonetheless, induction would be the way to go here.

3. ## Sum of fourth powers: proof by induction

Hello rajeev000 -

Originally Posted by rajeev000
Prove that the sum of fourth powers of the first n integers is 1/30n(n+1) (2n+1)(3n2+3n+1)
The formula you're trying to prove is:

$\sum_{i=1}^n n^4=\frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$

Prove by induction. Assume that this is true for some $n$. Then:

$\sum_{i=1}^{n+1} n^4=\frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1) + (n+1)^4$

When you're faced with an expression like this, don't multiply out all the brackets. Look for common factors. Here we have a common factor $(n+1)$. So factorise:

$\sum_{i=1}^{n+1} n^4=\frac{1}{30}(n+1)(n(2n+1)(3n^2+3n-1)+30(n+1)^3)$

Then you can multiply out within the long bracket, and simplify. You will then get
(I'll leave you to check this out):

$\sum_{i=1}^{n+1} n^4=\frac{1}{30}(n+1)(64n^4+39n^3+91n^2+89n+30)$

Don't be afraid of it! You know that we're looking for an expression which is the equivalent of the original one, with
$n$ replaced by $(n+1)$. So we should have a factor now of $(n+2)$ and another one of $(2n+3)$. So take out these factors, and you get (again, I'll leave you to supply the missing lines):

$\sum_{i=1}^{n+1} n^4=\frac{1}{30}(n+1)(n+2)(2n+3)(3n^2+9n+5)$

All that remains is to show that the final bracket is the same as $(3(n+1)^2+3(n+1)-1)$ - which it is - and to show that the formula is true when $n=1$. Can you do this now?

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# mathematical induction n to the fourth power

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