Mathematical Induction:n= n/6 (n+1) (n+2)

[cyberspace]

Prove that:
n = n/6 (n+1) (n+2)

Here are my steps:
n=k
- k = k/6 (k+1) (k+2)
n=k+1
- k + k+1 = (k+1)/6 (k+2) (k+3)
- k/6 (k+1) (k+2)+ k+1= (k+1)/6 (k+2) (k+3) <---This is where I got stuck. I tried factoring out 1/6 and k+1, but it still didn't equal (k+1)/6 (k+2) (k+3) .

How do I complete the last step? Thanks for your help in advance.

 

[tkhunny]

1) This makes no sense. Is there a summation somewhere?

2) You probably should do the first step before proceeding to the last step. Demonstrating continuation is of no value if it doesn't actually start somewhere.

 

[galactus]

It would appear the poster wants to prove by induction that \(\displaystyle \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}\)

You must've copied the problem down wrong, cyberspace, because what you have is not going to work because it is not true. That should be n^2, not n.

\(\displaystyle 1^{2}+2^{2}+3^{2}+4^{2}+...........+n^{2}=\frac{n(n+1)(2n+1)}{6}\)

I will skip the n=1 and induction steps and hop right to adding \(\displaystyle (n+1)^{2}\) to both sides.

\(\displaystyle 1^{2}+2^{2}+3^{2}+4^{2}+.....+n^{2}+(n+1)^{2}=\frac{n(n+1)(2n+1)}{6}+(n+1)^{2}\)

Now, If you factor the right side you get:

\(\displaystyle \frac{(n+1)(n+2)(2n+3)}{6}\)

Which is what you are trying to show. I will leave you plod through the algebra.