Telescoping SeriesDate: 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/ |
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