Evaluate these summations.

a) N(E)k=1 (k-k^3)

b) N-1(E)k=1 (k^2/N^2)

Note that (E) is the summation symbol.

Not sure how to do these. Help would be appreciated :)

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- May 7th 2011, 12:20 AMbrumby_3Summation
Evaluate these summations.

a) N(E)k=1 (k-k^3)

b) N-1(E)k=1 (k^2/N^2)

Note that (E) is the summation symbol.

Not sure how to do these. Help would be appreciated :) - May 7th 2011, 03:26 AMPlato
You need to study

*SPECIAL SUMS* - May 7th 2011, 04:32 AMtopsquark
- May 7th 2011, 05:41 AMSoroban
Hello, brumby_3!

These can be solved with basic algebra, but the process is long and tedious.

Don't know anyone who'd want to do it . . . Okay, it's me.

Quote:

$\displaystyle \displaystyle(a)\;\sum^n_{k=1}(k - k^3) $

$\displaystyle \text{The terms are: }\;\begin{Bmatrix}a_1 &=& 0 \\ a_2 &=& \text{-}6 \\ a_3 &=& \text{-}24 \\ a_4 &=& \text{-}60 \\a_5 &=& \text{-}120 \\ a_6 &=& \text{-}210 \end{Bmatrix}$ . . $\displaystyle \text{The partial sums are: }\;\begin{Bmatrix}S_1 &=& 0 \\ S_2 &=& \text{-}6 \\ S_3 &=& \text{-}30 \\ S_4 &=& \text{-}90 \\ S_5 &=& \text{-}210 \\ S_6 &=& \text{-}420 \end{Bmatrix}$

Take the differences of consecutive terms in the Partial Sums sequence.

Then take the differences of the differences . . . and so on.

$\displaystyle \begin{array}{ccccccccccccc} \text{Sequence} & 0 && \text{-}6 && \text{-}30 && \text{-}90 && \text{-}210 && \text{-}420 \\ \text{1st diff} && \text{-}6 && \text{-}24 && \text{-}60 && \text{-}120 && \text{-}210 \\ \text{2nd diff} &&& \text{-}18 && \text{-}36 && \text{-}60 && \text{-}90 \\ \text{3rd diff} &&&& \text{-}18 && \text{-}24 && \text{-}30 \\ \text{4th diff} &&&&& \text{-}6 && \text{-}6 \end{array}$

We see that the**4th**differences are constant.

This tells us that the generating function is of the**4th**degree.

The general quartic function is: .$\displaystyle S(n) \:=\:an^4 + bn^3 + cn^2 + dn + e$

Use the first 5 terms of the sequence and set up this system:

$\displaystyle \begin{array}{cccccc} S(1) \:=\:0\!: & a + b + c + d + e &=& 0 \\ S(2) \:=\:\text{-}6\!: & 16a + 8b + 4c + 2d + e &=& \text{-}6 \\ S(3) \:=\:\text{-}30\!: & 81a + 27b + 9c + 3d + e &=& \text{-}30 \\ S(4) \:=\:\text{-}90\!: & 256a + 64b + 16c + 4d + e &=& \text{-}90 \\ S(5) \:=\:\text{-}210\!: & 625a + 125b + 25c + 5df + e &=& \text{-}210 \end{array}$

Solve the system: .$\displaystyle a = \text{-}\tfrac{1}{4},\;b = \text{-}\tfrac{1}{2},\;c = \tfrac{1}{4},\;d = \tfrac{1}{2},\;e = 0 $

. . Hence: .$\displaystyle S(n) \;=\;\text{-}\tfrac{1}{4}n^4 - \tfrac{1}{2}n^3 + \tfrac{1}{4}n^2 + \tfrac{1}{2}n $

This can be drastically simplified . . .

$\displaystyle S(n) \;=\;\text{-}\tfrac{1}{4}n\left(n^3 + 2n^2 - n - 2\right) \;=\;\text{-}\tfrac{1}{4}n\bigg[n^2(n+2) - (n+2)\bigg] $

. . . . $\displaystyle =\;\text{-}\tfrac{1}{4}n(n+2)(n^2-1) \;=\;\text{-}\tfrac{1}{4}n(n+2)(n-1)(n+1)$

$\displaystyle \text{Therefore: }\;S(n) \;=\;-\frac{(n-1)n(n+1)(n+2)}{4} $

- May 7th 2011, 05:57 AMProve It
http://i22.photobucket.com/albums/b3...ofintegers.jpg

http://i22.photobucket.com/albums/b3...mofsquares.jpg

http://i22.photobucket.com/albums/b3...sumofcubes.jpg

From these diagrams it should be clear that

http://quicklatex.com/cache3/ql_4d49...2f0548e_l3.png

See if you can evaluate your sums now... - May 7th 2011, 06:05 AMtopsquark
The general idea was for the OP to look them up him/herself....

-Dan - May 7th 2011, 06:23 AMProve It