Summation

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May 7th 2011, 12:20 AM
brumby_3

Summation

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 AM
Plato

Quote:

Originally Posted by brumby_3

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 :)

You need to study SPECIAL SUMS
May 7th 2011, 04:32 AM
topsquark

Quote:

Originally Posted by brumby_3

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 :)

Specifically, what are
$\sum_{k = 1}^N k$

$\sum_{k = 1}^N k^2$

and
$\sum_{k = 1}^N k^3$

-Dan
May 7th 2011, 05:41 AM
Soroban

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:

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

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

Take the differences of consecutive terms in the Partial Sums sequence.
Then take the differences of the differences . . . and so on.

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

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: . $S(n) \:=\:an^4 + bn^3 + cn^2 + dn + e$

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

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

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

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

This can be drastically simplified . . .
$S(n) \;=\;\text{-}\tfrac{1}{4}n\left(n^3 + 2n^2 - n - 2\right) \;=\;\text{-}\tfrac{1}{4}n\bigg[n^2(n+2) - (n+2)\bigg]$

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

$\text{Therefore: }\;S(n) \;=\;-\frac{(n-1)n(n+1)(n+2)}{4}$
May 7th 2011, 05:57 AM
Prove 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 AM
topsquark

The general idea was for the OP to look them up him/herself....

-Dan
May 7th 2011, 06:23 AM
Prove It

Quote:

Originally Posted by topsquark

The general idea was for the OP to look them up him/herself....

-Dan

It might have been hard for the OP to find these particular proofs, seeing as they're from an ebook I bought... Besides, there's still plenty of work for the OP to do.