Prove: each square is the sum of 2 consecutive triangular numbers

**courteous** · October 1st 2010, 01:24 AM

Is the following an acceptable proof (be harsh

)?

$t_k=1+2+3+\ldots+k=\frac{(k+1)k}{2}$
$t_{k-1}=1+2+3+\ldots+(k-1)=\frac{k(k-1)}{2}$
$\Longrightarrow t_{k-1}+t_k=k^2 \text{ } \square ?$

I mean, is the induction also required? Even if not, I need help with it:

Base case - $k=1$ : $1^2=t_0+t_1=0+1=1$ ; define $t_0=0$ .

Induction - $k=k$ : suppose $k^2=t_{k-1}+t_k$ .
Then $(k+1)^2=t_k+t_{k+1}=?$ ... how do I use the supposition?

**Prove It** · October 1st 2010, 01:50 AM

Just show that

$\frac{(k + 1)k}{2} + \frac{(k - 1)k}{2} = k^2$ ...

$\frac{(k + 1)k}{2} + \frac{(k - 1)k}{2} = \frac{k^2 + k}{2} + \frac{k^2 - k}{2}$

$= \frac{k^2 + k + k^2 - k}{2}$

$= \frac{2k^2}{2}$

$= k^2$ .

**courteous** · October 1st 2010, 02:16 AM

Done that, just didn't write it down.

The Halmos $\square$ sign (with a "?" mark) indicates whether this is enough ... or is induction also needed (even if it is not, still help me with the induction step

).

MOD request: Can you please restore my spam-hijacked question?

**Prove It** · October 1st 2010, 02:21 AM

$t_{k + 1} + t_k = \frac{(k + 2)(k + 1)}{2} + \frac{(k+1)k}{2}$

$= \frac{k^2 + 3k + 2}{2} + \frac{k^2 + k}{2}$

$= \frac{2k^2 + 4k + 2}{2}$

$= \frac{2(k^2 + 2k + 1)}{2}$

$= k^2 + 2k + 1$

$= (k + 1)^2$ .

**courteous** · October 1st 2010, 02:33 AM

Is this induction (not saying that it isn't, I just don't know)? I mean, isn't it necessary to use $k^2=t_{k-1}+t_k$ equality?

**Bruno J.** · October 2nd 2010, 10:51 AM

Originally Posted by courteous

Is the following an acceptable proof (be harsh

)?

$t_k=1+2+3+\ldots+k=\frac{(k+1)k}{2}$
$t_{k-1}=1+2+3+\ldots+(k-1)=\frac{k(k-1)}{2}$
$\Longrightarrow t_{k-1}+t_k=k^2 \text{ } \square ?$

I mean, is the induction also required? Even if not, I need help with it:

Base case - $k=1$ : $1^2=t_0+t_1=0+1=1$ ; define $t_0=0$ .

Induction - $k=k$ : suppose $k^2=t_{k-1}+t_k$ .
Then $(k+1)^2=t_k+t_{k+1}=?$ ... how do I use the supposition?

In my opinion, even a drawing should suffice in this case! Take a square like this

* * * *
* * * *
* * * *
* * * *

and split it up into two triangles along the main diagonal, with one of the triangles including the diagonal.

**courteous** · October 2nd 2010, 12:13 PM

This is just a special case of $t_3+t_4$ .

Guess I learned something about proving things ... and not trying to prove the obvious.

**Archie Meade** · October 2nd 2010, 02:13 PM

Originally Posted by courteous

Done that, just didn't write it down.

The Halmos $\square$ sign (with a "?" mark) indicates whether this is enough ... or is induction also needed (even if it is not, still help me with the induction step

).

MOD request: Can you please restore my spam-hijacked question?

You have given a very nice proof...

Suppose you were unaware of the proof you gave in your first post and were feeling lazy...

Here is the "Proof By Induction" method

The triangular numbers are 1, 3, 6, 10, 15, 21, 28,....

1=1^2

1+3=4=2^2

3+6=9=3^2 etc

P(k)

$t_{k-1}+t_k=k^2$

P(k+1)

$t_k+t_{k+1}=(k+1)^2$

Try to show that P(k) being valid (even if we don't know whether it is or not)
will cause P(k+1) to be valid.

Hence write P(k+1) in terms of P(k).

Proof

$t_k+t_{k+1}=t_k+t_{k-1}+k+(k+1)$

$=k^2+2k+1$ if P(k) really is valid

$=(k+1)^2$

Now the line of dominoes is in place, hence you need to check if the first one falls.

$1+3=4=2^2$ true

$0+1=1=1^2$ true, depending on where you want to start from.

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