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

• Oct 1st 2010, 01:24 AM
courteous
Prove: each square is the sum of 2 consecutive triangular numbers
Is the following an acceptable proof (be harsh (Nod))?

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

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

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

• Induction - $\displaystyle k=k$: suppose $\displaystyle k^2=t_{k-1}+t_k$.
Then $\displaystyle (k+1)^2=t_k+t_{k+1}=?$ ... how do I use the supposition? (Blush)
• Oct 1st 2010, 01:50 AM
Prove It
Just show that

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

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

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

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

$\displaystyle = k^2$.
• Oct 1st 2010, 02:16 AM
courteous
Done that, just didn't write it down. (Wink) The Halmos $\displaystyle \square$ 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 (Blush)).

MOD request: Can you please restore my spam-hijacked question?
• Oct 1st 2010, 02:21 AM
Prove It
$\displaystyle t_{k + 1} + t_k = \frac{(k + 2)(k + 1)}{2} + \frac{(k+1)k}{2}$

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

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

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

$\displaystyle = k^2 + 2k + 1$

$\displaystyle = (k + 1)^2$.
• Oct 1st 2010, 02:33 AM
courteous
Is this induction (not saying that it isn't, I just don't know)? I mean, isn't it necessary to use $\displaystyle k^2=t_{k-1}+t_k$ equality?
• Oct 2nd 2010, 10:51 AM
Bruno J.
Quote:

Originally Posted by courteous
Is the following an acceptable proof (be harsh (Nod))?

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

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

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

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

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.
• Oct 2nd 2010, 12:13 PM
courteous
This is just a special case of $\displaystyle t_3+t_4$. (Smirk)
Guess I learned something about proving things ... and not trying to prove the obvious. (Happy)
• Oct 2nd 2010, 02:13 PM
Quote:

Originally Posted by courteous
Done that, just didn't write it down. (Wink) The Halmos $\displaystyle \square$ 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 (Blush)).

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)

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

P(k+1)

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

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

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

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

$\displaystyle =(k+1)^2$

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

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

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