Summation formula without a variable specified

Question

I'm attempting to solve this problem, and I've reached an impasse. Because I'm attempting to solve for the first value.

$$ \sum_{i=1}^N N^3 = \left( \sum_{i=1}^N i \right)^2 $$

I then get $(N(N+1)/2)^2$ which then expands to $(N^4 + 2N^3 + N^2)/2$.

After this point I'm lost and not sure where to go to continue the problem.

score 0 · Accepted Answer · answered Sep 04 '18 at 02:43

0

Guide:

try mathematical induction.

Base case: left as exercise.

Suppose $$\sum_{i=1}^{N-1} i^3 = \left( \sum_{i=1}^{N-1} i\right)^2$$

You want to show that

$$\sum_{i=1}^{N} i^3 = \left( \sum_{i=1}^{N} i\right)^2=\left(\frac{N(N+1)}{2} \right)^2$$

Start from LHS,

$$\sum_{i=1}^{N} i^3 = \sum_{i=1}^{N-1} i^3 + N^3=\left( \sum_{i=1}^{N-1} i\right)^2 + N^3=\left(\frac{(N-1)N}{2} \right)^2+N^3$$

Try to show that $\left(\frac{(N-1)N}{2} \right)^2+N^3=\left(\frac{N(N+1)}{2} \right)^2$

btw, there is a typo in your question, the LHS should be $\sum_{i=1}^{N} i^3$

answered Sep 04 '18 at 02:43

Siong Thye Goh

The question I was given did have $\sum\limits_{i=1}^N N^3$ for LHS. I though it could be slightly odd, but It was what I was given. – kyle newton Sep 04 '18 at 03:07
1

I see, most likely a typo, let $N=2$ and we can see that it is not true. – Siong Thye Goh Sep 04 '18 at 03:09

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