# Math Question, need help badly.

1. Mar 7, 2005

### lvlastermind

I've been stuck on this question for awhile.

Q: Square numbers 1, 4, 9, 16, 25.... are the values of the function s(n)=n^2, when n is a positive integer. The triangular numbers t(n)=(n(n+1))/2 are the numbers t(1)=1, t(2)=3, t(3)=6, t(4)=10.

Prove: For all positive integers n, s(n+1) = t(n) + t(n+1)

I've tride alot of things and come to the conclusion that I cant get my answer by using polynomials. I think that if you subsitiute t(n)=(n(n+1))/2 into the equation and simplify to get (n+1)^2 I will be done. My problem is that I'm having troubles doing this. Any sugestions???

2. Mar 7, 2005

### TsunamiJoe

$$( \frac{1} {2} ( n - 1 ) n ) ^ 2$$ is most similiar to what your talking about, personaly i prefer the: $$( \frac{n} {2} ( n + 1 ) ) ^ 2$$

orr this could also be it: $$S_N = \frac{N}{2} ( A_1 + A_N)$$

and again im only regurgitating equations on you that look like what you could be searching for

Last edited: Mar 7, 2005
3. Mar 7, 2005

### vincentchan

expand the right hand side and rearrange it into the form of (n+1)^2 = s(n+1)

note:
RHS = t(n)+t(n+1) = n(n+1)/2 + (n+1)((n+1)+1)/2

4. Mar 7, 2005

### vitaly

After changing n to (n+1), you get:

(n+1)^2 = n(n+1)/2 + (n+1)((n+1)+1)/2)

= (n+1)^2 = (n^2+n)/2 + (n+1)(n+2)/2

= (n+1)^2 = (n^2+n)/2 + (n^2+3n+2)/2

Since they have common denominators, we can add the right side together:

= (n+1)^2 = (2n^2 + 4n + 2)/2

= (n+1)^2 = (2n+2)(n+1)/2

= (n+1)^2 = 2(n+1)(n+1)/2

The two's cancel out, which gives the needed proof:

= (n+1)^2 = (n+1)(n+1)

5. Mar 7, 2005

### lvlastermind

how did you go from
= (n+1)^2 = (2n^2 + 4n + 2)/2

to

= (n+1)^2 = (2n+2)(n+1)/2

6. Mar 7, 2005

### lvlastermind

Thanks for the help all, I got it.