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Given the sequence:

$\begin{gather*} 1^2 \\\ 2^2 - 1^2 \\\ 3^2 - 2^2 + 1^2\\\ 4^2 - 3^2 + 2^2 - 1^2\\\ 5^2 - 4^2 + 3^2 - 2^2 + 1^2\\\ \vdots \end{gather*}$

It's clear that the pattern results in the triangular number sequence $(1,3,6,10,15,\ldots)$ but I'm having trouble putting it into a formula in terms of $n$. Any tips are appreciated!

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    @Zev - I have that page open, and read it before posting my question, but don't see that representation of the sequence in there. But you answered my question. Thanks! Also, thanks for the formatting. I'll figure that out before I post again.2011-07-24

5 Answers 5

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For each pair of squares:

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

From which the result follows for an even number of terms. For an odd number of terms add $-0^2$ to the end of the sequence, to get an even number of terms with the same sum.

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$T_n = \frac{n(n+1)}{2}$ and $T_{n-1} = \frac{n(n-1)}{2}$ so $T_n+T_{n-1} = n^2$, i.e. $T_n= n^2- T_{n-1}$.

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So knowing $T_1=1$, you can use induction to show your result.


Another way of looking at the question is to see the triangle number as the sum of L-shaped strips and each L-shape as the difference between consecutive squares.

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    +1! In "A Brief History of Time", Hawking says that he did not use formulas in his book because "the number of readers is inversely proportional to the number of formulas", or something along that line. :)2011-07-25
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I'm not sure I answered your question; I was only making sure I understood what you are asking. However, I will post a hint: A good way to prove something like $\sum_{k=1}^n (-1)^{n-k}k^2=T_n$ is via induction on $n$. If you can prove the above statement is true when $n=1$, and you can prove that, whenever it is true for $n=a$, it will also be true for $n=a+1$, then it will be true for all $n$.

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    Ah, I see! I'm glad to have helped!2011-07-24
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Yet another way to prove it is by first showing that $n^2=\binom{n+1}{2}+\binom{n}2,$ and then using a form of telescoping: $\binom {n+1}2+\binom n2-\binom n2-\binom{n-1}2+\cdots+(-1)^n\binom 22-(-1)^n\binom 22=\binom{n+1}2.$

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    Thanks to both of you!2011-07-25
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For $n=1,2,3,\dots$, let $T_n$ be the $n$-th triangular number, and let $U_n$ be the $n$-th element of our sequence.

We can use the following principle, which we deliberately state a little vaguely. If two sequences begin in the same way, and satisfy the same recurrence, then the sequences are the same. (If the recurrence is second-order, as in the Fibonacci recurrence, "start in the same way" means that the first two terms agree.)

What recurrence shall we use? Note that $U_1=T_1$, so the two sequences start in the same way. It is obvious that $U_{n-1}+U_n=n^2$, or equivalently $U_n=n^2-T_{n-1}$. So if we can show that $T_n=n^2-T_{n-1}$, we will be finished.

In this case, we know an explicit formula for the $n$-th triangular number, since, essentially from the definition of $T_n$, we know that $T_n=1+2+\cdots+n$, and the sum of this arithmetic series is well-known, it is $n(n+1)/2$. Thus $T_n+T_{n-1}=\frac{(n)(n+1)}{2}+\frac{(n-1)(n)}{2}=n^2,$ and therefore the sequence $(T_n)$ satisfies the recurrence $T_n=n^2-T_{n-1}$. This completes the proof.

The general idea is quite useful, since in combinatorics, explicit formulas are (after a while) relatively uncommon, but recurrences are quite common.