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### Telescoping Series

```
Date: 03/28/2001 at 16:21:49
From: KSuchanek
Subject: Re: Number patterns

I am trying to find the nth equation in the triangular number formula.

This is for even numbers:

2 = 1-3 (-2)
4 = 1-3+6-10 (-6)
6 = 1-3+6-10+15-21 (-12)

Can you help me with the formula for these patterns?

Thank you,
Karen Suchanek
```

```
Date: 03/30/2001 at 11:26:24
From: Doctor Peterson
Subject: Re: Number patterns

Hi, Karen.

Let's first improve your notation; the equal signs are confusing,
since you aren't really saying that 2 = 1-3, and the pattern involves
not equations but sequences or series. I'd say it this way, using s[n]
to mean the nth term in your sequence, that is, the alternating sum of
the first n triangular numbers:

s[1] = 1 = 1
s[3] = 1-3+6 = 4
s[5] = 1-3+6-10+15 = 9
s[7] = 1-3+6-10+15-21+34 = 16

and

s[2] = 1-3 = -2
s[4] = 1-3+6-10 = -6
s[6] = 1-3+6-10+15-21 = -12

Each case can be handled as a telescoping series. Using t[n] for the
nth triangular number, we can write

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

and then

s[n] = t[1] - t[2] + ... +- t[n]

where the signs alternate.

Notice first that

t[n+1] - t[n] = (1+2+...+n+(n+1)) - (1+2+...+n) = n+1

For odd n, your series can be paired off (except for the first term)
this way:

s[2k+1] = t[1] + (-t[2] + t[3]) + ... + (-t[2k] + t[2k+1])
= 1 + 3 + ... + (2k+1)
= (k+1)^2

using the well-known formula for the sum of odd numbers.

For even n, we can pair off all terms:

s[2k] = (t[1] - t[2]) + (t[3] - t[4]) + ... + (t[2k-1] - t[2k])
= -2 + -4 + ... + -2k
= -2(1 + 2 + ... + k)
= -2t[k]
= -k(k+1)

We say that the series "telescopes" because successive terms "slide
together" to make a simpler series.

Let me know if you need more help with this.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series

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