Functions, Statistics, and Trigonometry problem

[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?

 

[honestrosewater]

Can you set up the equation? I get
$$(n + 1)^{2} = \frac{[n(n + 1)] + [(n + 1)(n + 2)]}{2}$$
Can you simplify that?

 

[lvlastermind]

Thats what my book has for the first step but I am confused about where you got the term (n+2) from putting t(n)=(n(n+1))/2 in.

 

[honestrosewater]

t(n) = (n(n + 1))/2
so t(n + 1) = [(n + 1)(n + 1 + 1)]/2.

 

[lvlastermind]

My book has the next step as t(n) + t(n+1)=(1/2)(n+1)(n+(n+2))

how did it get there

 

[honestrosewater]

I have no idea. Of course, since [a/b = a(1/b)] you can see how they can multiply by 1/2 instead of dividing by 2. I don't know about the rest. The next step after
$$(n + 1)^{2} = \frac{[n(n + 1)] + [(n + 1)(n + 2)]}{2}$$
is
$$(n + 1)^{2} = \frac{[n^{2} + n] + [n^{2} + 3n + 2]}{2}$$
Can you take it from there?
Edit: Do you know how to simplify (n +1)(n + 2)?

 

[lvlastermind]

Yeah I can, thanks a lot honestrosewater

 

[honestrosewater]

Just curious- does anyone else see how they got
t(n) + t(n+1)=(1/2)(n+1)(n+(n+2))?

 

[dextercioby]

Beside the very obvious

$$t(n)=\frac{n(n+1)}{2} \Rightarrow t(n+1)=\frac{(n+1)(n+2)}{2}$$

Therefore the sum is

$$t(n)+t(n+1)=\frac{(n+1)}{2}[n+(n+2)]$$

Daniel.

P.S.Is there any other simpler way...?

 

[honestrosewater]

The not-skipping-steps-in-between way of adding them is
$$\frac{[n(n + 1)]}{2} + \frac{[(n + 1)(n + 2)]}{2}$$
What are the actual steps in getting to
$$\frac{(n+1)}{2}[n+(n+2)]$$?
Or at least some of the steps? I don't see them.

 

[dextercioby]

How about factoring
$$\frac{n+1}{2}$$ ?

It's in both terms.

Daniel.

 

[honestrosewater]

... thanks.