Prove that the sum of fourth powers of the first n integers is 1/30n(n+1) (2n+1)(3n2+3n+1)
Hello rajeev000 -
The formula you're trying to prove is:
$\displaystyle \sum_{i=1}^n n^4=\frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$
Prove by induction. Assume that this is true for some $\displaystyle n$. Then:
$\displaystyle \sum_{i=1}^{n+1} n^4=\frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1) + (n+1)^4$
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 $\displaystyle (n+1)$. So factorise:
$\displaystyle \sum_{i=1}^{n+1} n^4=\frac{1}{30}(n+1)(n(2n+1)(3n^2+3n-1)+30(n+1)^3)$
Then you can multiply out within the long bracket, and simplify. You will then get (I'll leave you to check this out):
$\displaystyle \sum_{i=1}^{n+1} n^4=\frac{1}{30}(n+1)(64n^4+39n^3+91n^2+89n+30)$
Don't be afraid of it! You know that we're looking for an expression which is the equivalent of the original one, with $\displaystyle n$ replaced by $\displaystyle (n+1)$. So we should have a factor now of $\displaystyle (n+2)$ and another one of $\displaystyle (2n+3)$. So take out these factors, and you get (again, I'll leave you to supply the missing lines):
$\displaystyle \sum_{i=1}^{n+1} n^4=\frac{1}{30}(n+1)(n+2)(2n+3)(3n^2+9n+5)$
All that remains is to show that the final bracket is the same as $\displaystyle (3(n+1)^2+3(n+1)-1)$ - which it is - and to show that the formula is true when $\displaystyle n=1$. Can you do this now?
Grandad