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 Math Help > Sequences and Series > Series > Polynomial Geometric

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A "polynomial geometric" series is one in which the kth term is the product of an m-degree polynomial in k times xk.

 n ∑ k=0 (a0 + a1k + a2k2 + ... + am-1km-1 + amkm) xk

The way to find such a sum is to telescope the m-degree polynomial by taking successive differences.  That is, by multiplying the series by (1-x), the resulting series is a polynomial geometric series of degree m-1.  The first and last terms, don't cancel, however.  Then, repeating the process, i.e. multiplying again by (1-x), the series eventually becomes purely geometric, with m non-canceling terms at the beginning, and another m non-canceling terms at the end.  Then, multiply by (1-x) one more time, as you normally do with a geometric series, to make "most" of the terms cancel completely.

It turns out that the non-canceling terms become quite a challenge, especially if n < 2m -- that is, when the non-canceling terms overlap.  For this reason, I will confine the rest of this article to the case when n approaches infinity, so the non-canceling terms will occur only at the beginning of the series.  This doesn't actually cause any loss of generality, though, because a finite series can be expressed as the difference of two infinite series with different starting points.

Let S be the sum of an infinite series of this form,

 S = ∞ ∑ k=0 (a0 + a1k + a2k2 + ... + am-1km-1 + amkm) xk

Now, multiply the sequence by (1-x)m+1, giving

S(1-x)m+1 =

 C(m+1,0) ∞∑ k=0 (a0 + a1k + a2k2 + ... + am-1km-1 + amkm) xk
 - C(m+1,1) ∞∑ k=1 (a0 + a1(k-1) + a2(k-1)2 + ... + am-1(k-1)m-1 + am(k-1)m) xk
 . . .
 + (-1)iC(m+1,i) ∞∑ k=i (a0 + a1(k-i) + a2(k-i)2 + ... + am-1(k-i)m-1 + am(k-i)m) xk
 . . .
 + (-1)m+1C(m+1,m+1) ∞∑ k=m (a0 + a1(k-m) + a2(k-m)2 + ... + am-1(k-m)m-1 + am(k-m)m) xk

The beauty of this telescoping method is that all the terms of S(1-x)m+1 except for the first m+1 terms cancel completely.  The only terms left are:

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The webmaster and author of this Math Help site is Graeme McRae.