My Math Forum summation of finite series (method of least differences)

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 November 5th, 2011, 08:05 AM #1 Newbie   Joined: Nov 2011 Posts: 2 Thanks: 0 summation of finite series (method of least differences) im having trouble with computing summation of finite series for series higher than a polynomial of the second order. the problem is even though i can get the equation finding the summation using the sigma n,n^2,n^3 formula is very cumbersome. besides the book seems to use some other method. tl,dr solve this please 1+2+29+130+377+866+1717+...
November 5th, 2011, 10:08 AM   #2
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Re: summation of finite series (method of least differences)

Hello, a_ntimage!

Quote:
 $\text{Find a formula for: }\:S(n) \:=\:1\,+\,2\,+\,29\,+\,130\,+\,377\,+\,866\,+\,17 17\,+\,\text{ . . . }$

We have the sequence:[color=beige] .[/color]$1,\,3,\,32,\,162,\,539,\,1405,\,322,\,\text{ . . .}$

Take the differences of consecutive terms,
[color=beige]. . [/color]then take differences of the differences, and so on.

$\begin{array}{ccccccccccccccc}\text{Sequence} && 1 && 3 && 32 && 162 && 539 && 1405 && 3122 \\ \\ \text{1st diff.} &&& 2 && 29 && 130 && 377 && 866 && 1717 \\ \\ \text{2nd diff.} &&&& 27 && 101 && 247 && 489 && 851 \\ \\ \text{3rd diff.} &&&&& 74 && 146 && 242 && 362 \\ \\ \text{4th diff.} &&&&&& 72 && 96 && 120 \\ \\ \text{5th diff.} &&&&&&& 24 && 24 \end{array}$

The fifth differences are constant.
[color=beige]. . [/color]Hence, the generating function is of the fifth degree.

The general fifth-degree function is:[color=beige] .[/color]$f(n) \:=\:an^5\,+\,bn^4\,+\,cn^3\,+\,dn^2\,+\,en\,+\,f$

Use the first six terms of the sequence and set up a system of equations:

$\begin{array}{ccccccc}f(1)= 1 &a\,+\,b\,+\,c\,+\,d\,+\,e\,+\,f=&1 \\ \\ f(2)= 3 &32a\,+\,16b\,+\,8c\,+\,4d\,+\,2e\,+\,f=&3 \\ \\ f(3)= 32 &243a\,+\,81b\,+\,27c\,+\,9d\,+\,3e\,+\,f=&32 \\ \\ f(4)= 162 &1024a\,+\,256b\,+\,64c\,+\,16d\,+\,4e\,+\,f=&162 \\ \\ f(5)= 539 &3125a\,+\,625b\,+\,125c\,+\,25d\,+\,5e\,+\,f=$

$\text{After a }lot\text{ of Algebra, we get: }\;\begin{Bmatrix} a=&\frac{1}{5}=&d=&-\frac{1}{2} \\ \\ \\ b=&e=&\frac{59}{30} \\ \\ \\ c=&-\frac{2}{3}=&f=&0 \end{Bmatrix}=$

$\text{Therefore: }\:f(n) \;=\;\frac{1}{5}n^5\,-\,\frac{2}{3}n^3 \,-\,\frac{1}{2}n^2\,+\,\frac{59}{30}n \;=\;\frac{n}{30}(6n^4\,-\,20n^2\,-\,15n\,+\,59)$

 November 5th, 2011, 08:44 PM #3 Newbie   Joined: Nov 2011 Posts: 2 Thanks: 0 Re: summation of finite series (method of least differences) ah thanks... but i figured out that bit.... the problem is how to sum this series with only my 3 formula i.e. sigma n,n^2,n^3? also my constants were wrong huh -___-

 Tags differences, finite, method, series, summation

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