How can I prove s(n+1) = t(n) + t(n+1) for all positive integers n?

[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 a lot of things and come to the conclusion that I can't 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?

 

[TsunamiJoe]

im not completely sure what your talking about, but i think your equation is wrong

$$( \frac{1} {2} ( n - 1 ) n ) ^ 2$$ is most similar 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 I am only regurgitating equations on you that look like what you could be searching for

 

[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

 

[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)

 

[lvlastermind]

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

to

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

 

[lvlastermind]

Thanks for the help all, I got it.