1. ## Sum of series

I've done the first part of the question but I'm not sure about the second part. I have some special results that might be required for it but they possibly aren't required.

2. Originally Posted by Stuck Man
I've done the first part of the question but I'm not sure about the second part. I have some special results that might be required for it but they possibly aren't required.
for the second part, use induction to prove that
do you know how is it done ?

3. Yes but I would usually have a series with an nth term.

4. Originally Posted by Stuck Man
Yes but I would usually have a series with an nth term.

$\displaystyle 1^3 +2^3+3^3+ ... + n^3 = \frac {n^2}{4} (n+1)^2$

$n=1$

$\displaystyle 1^3 = \frac {1^2}{4} ( 1+1) ^2 \Rightarrow 1=1$

(show for n=2, n=3 )

assume that it's true for n=k :

$\displaystyle 1^3 +2^3+3^3+ ... + k^3 = \frac {k^2}{4} (k+1)^2$

now with assumption that is true for n=k , show that is true for n=k+1 ....

$\displaystyle 1^3 +2^3+3^3+ ... + k^3 +(k+1)^3= \frac {(k+1)^2}{4} ((k+1)+1)^2$

can you do that ?

hint : $\displaystyle 1^3 +2^3+3^3+ ... + k^3 = \frac {k^2}{4} (k+1)^2$ so add $\displaystyle(k+1)^3$ to $\frac {k^2}{4} (k+1)^2$ and you should get the $\displaystyle \frac {(k+1)^2}{4} ((k+1)+1)^2$

5. I am able to do proof by induction like that. I don't think this is answering the question using the identity given.

6. This is not the required proof. I have a book that has a similar proof to that required starting from an identity for r^2. This thread was on the second page and would not get any replies now if I weren't putting it back up to the top.

7. Originally Posted by Stuck Man
This is not the required proof. I have a book that has a similar proof to that required starting from an identity for r^2. This thread was on the second page and would not get any replies now if I weren't putting it back up to the top.
hmmm.... okay

did you try to do it with that identity ?

$\displaystyle (1\cdot 2 \cdot 3 - 3\cdot 1 \cdot 2 +1 )^3 + ( 2\cdot 3 \cdot 4 - 3\cdot 2 \cdot 3 +1)^3 + ... + (n\cdot (n+1) \cdot (n+2) - 3n \cdot (n+1) +n)^3 = \frac {n^2}{4} \cdot (n+1)^2$

and do induction

8. It will still be a proof by induction. I am going to scan a page from a book that shows the kind of answer required.

9. I've attached a scan with the method.