I was looking over this problem and im pretty sure you use induction.. but what would your base case be and when subing (k+1) for n, what is the inital formula we want to achieve? I dont exactly know how to approach this problem.

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- July 13th 2010, 07:45 AMjess0517Another proof by induction.
I was looking over this problem and im pretty sure you use induction.. but what would your base case be and when subing (k+1) for n, what is the inital formula we want to achieve? I dont exactly know how to approach this problem.

- July 13th 2010, 07:52 AMchiph588@
- July 13th 2010, 09:42 AMGrandad
Hello jess0517

We have to prove that is an integer for all .

First we factorise:

So we must prove that is a multiple of 6; in other words, that it is a multiple of 2 and of 3.

First, note that one of and is even. So is even.

Next, note that is either a multiple of 3, in which case there is nothing left to prove; or it leaves a remainder 1 or 2 when divided by 3.

Case I: leaves a remainder 1; in which case . Therefore

Case II: leaves a remainder 2; in which case . Therefore

This concludes the proof.

Grandad

- July 13th 2010, 09:43 AMjess0517
- July 13th 2010, 12:03 PMRenji Rodrigo

summation of integer( k^2) is integer - July 13th 2010, 01:37 PMArchie Meade
**P(k)**

is an integer ?

**P(k+1)**

is also an integer ?

Try to prove that P(k) being true**causes**P(k+1) to be true.

**Proof**

If k is an integer, so is k+1 and (k+1)(k+1).

Hence, if P(k) is true, it causes P(k+1) to be true.

All that remains is to test the proposition P(k) for the first integer.