Help on Induction: Proving \sum_{r=1}^n (r^5)

[PhY_InTelLecT]

How do i prove this $$\sum_{r=1}^n (r^5)=1/12n^2(n+1)^2(2n^2+2n-1)$$

when i have already proven this 2.. That $$\sum_{r=1}^n (r^3)=1/4n^2(n+1)^2 <br />$$ and $$\sum_{r=1}^n (r^3+3r^5)=1/2n^3(n+1)^3$$

The problem is that i know how to solve the first 2 but dun know how to solve the big one using the other 2.

 

[VietDao29]

So you've proven that:
$$\sum_{r = 1} ^ {n} (r ^ 3 + 3r ^ 5) = \frac{1}{2}n ^ {3} (n + 1) ^ {3}$$ and $$\sum_{r = 1} ^ {n} r ^ 3 = \frac{1}{4}n ^ {2} (n + 1) ^ {2}$$.
To find: $$\sum_{r = 1} ^ {n} r ^ 5$$
You should note that:
$$\sum_{r = 1} ^ {n} (r ^ 3 + 3r ^ 5) \ = \ \sum_{r = 1} ^ {n} r ^ 3 \ + \ \sum_{r = 1} ^ {n} 3r ^ 5 = \sum_{r = 1} ^ {n} r ^ 3 \ + \ 3\sum_{r = 1} ^ {n} r ^ 5$$.
Rearrange it a bit, we have:
$$\sum_{r = 1} ^ {n} r ^ 5 = \frac{1}{3} \times \left( \sum_{r = 1} ^ {n} (r ^ 3 + 3r ^ 5) - \sum_{r = 1} ^ {n} r ^ 3 \right)$$.
Can you go from here?