Simplifying summation

**borracho** · March 10th 2009, 06:46 PM

Hello,

I have a summation from k = 1,..., n of (m-k+1)(n-k+1), and I would like to simplify this in terms of just m and n.. I know that I need to write out the sum first, so

sum = (mn) + (m-1)(n-1) + (m-2)(n-2) + ... + (m-n+1)(1),
(is this right?)

but I'm having trouble proceeding from this to the next step. Can anyone help?

**Soroban** · March 10th 2009, 07:11 PM

Hello, borracho!

I would use these summation formulas:

. . $\sum^n_{k=1}k \;=\;\frac{k(k+1)}{2}$

. . $\sum^n_{k=1}k^2 \:=\:\frac{k(k+1)(2k+1)}{6}$

$S \;=\;\sum^m_{k=1}(m-k+1)(n-k+1)$

We have: . $\bigg[(m+1)-k\bigg]\,\bigg[(n+1)-k\bigg] \;=\;(m+1)(n+1) - (m+n+2)k + k^2$

Then: . $S \;\;=\;\;\sum^n_{k=1}\bigg[(m+1)(n+1) - (m+n+2)k + k^2\bigg]$

. . . . . $S \;\;=\;\;(m+1)(n+1)\sum^n_{k=1}1 \;-\; (m+n+2)\sum^n_{k=1} k \;+\; \sum^n_{k=1}k^2$

. . . . . $S \;\;=\;\;(m+1)(n+1)\cdot n \;-\;(m+n+2)\cdot\frac{n(n+1)}{2} \;+\;\frac{n(n+1)(2n+1)}{6}$

I'll let you simplify it . . .

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