# Double Summation (multiplying)

• Feb 28th 2010, 01:21 PM
krazyxazn
Double Summation (multiplying)
http://img202.imageshack.us/img202/8954/capturemu.jpg

Is there some kind of formula/shortcut for solving something like this? I could write it all out, but seems unpractical.

What I'm currently using:
[ 50(50+1)/2 - ( i - 1)(i -1 +1)/2 ] * i
Just fill out that 50 times :(.

I made a program to solve it and it gave out: 834275.
• Feb 28th 2010, 07:13 PM
Soroban
Hello, krazyxazn!

We're expected to evaluate it without cranking out the sums,
but it still takes some knowledge and quite a bit of algebra.

Quote:

$S \;=\;\sum^{50}_{i=1} \sum^{50}_{j=i} ij$

The inner sum is:

. . $\sum^{50}_{j=1} ij \;=\;i(i) + i(i+1) + i(i+2) + i(i+3) + \hdots + i(50)$

. . . . . . $=\;i \underbrace{\bigg[i + (i+1) + (i+2) + (i+3) + \hdots + 50\bigg]}_{T = \text{ sum of integers from }i\text{ to } 50}$

. . $T \;=\;\frac{50\cdot51}{2} - \frac{(i-1)i}{2} \;=\;\frac{2550+i-i^2}{2}$

Hence, the inner sum is: . $\sum^{50}_{j=i} ij \;=\;\left(\frac{2550+i - i^2}{2}\right) i \;=\;\frac{1}{2}(2550i + i^2 - i^3)$

The entire sum is: . $S \;=\;\sum^{50}_{i=1}\frac{1}{2}(2550i + i^2 - i^3)$

And we have: . $S \;=\;1275\sum^{50}_{i=1} i \;+\;\frac{1}{2}\sum^{50}_{i=1} i^2 \;-\;\frac{1}{2}\sum^{50}_{i=1} i^3$

We are expected to know the formulas for these three sums. . **

. . $S \;=\;1275\cdot\frac{50\cdot51}{2} \;+\;\frac{1}{2}\cdot\frac{50(51)(101)}{6} \;-\; \frac{1}{2}\cdot\left(\frac{50\cdot51}{2}\right)^2$

Therefore: . $\boxed{S \;=\;834,\!275}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** . Formulas

. . $\sum^n_{i=1}i \;\;\,=\quad 1+2+3+\hdots + n \quad =\;\; \frac{n(n+1)}{2}$

. . $\sum^n_{i=1} i^2 \;=\; 1^2+2^2+3^2+\hdots + n^2 \;=\; \frac{n(n+1)(2n+1)}{6}$

. . $\sum^n_{i=1} i^3 \;=\; 1^3 + 2^3 + 3^3 + \hdots + n^3 \;=\; \left[\frac{n(n+1)}{2}\right]^2$

• Feb 28th 2010, 08:10 PM
Drexel28
Quote:

Originally Posted by krazyxazn
http://img202.imageshack.us/img202/8954/capturemu.jpg

Is there some kind of formula/shortcut for solving something like this? I could write it all out, but seems unpractical.

What I'm currently using:
[ 50(50+1)/2 - ( i - 1)(i -1 +1)/2 ] * i
Just fill out that 50 times :(.

I made a program to solve it and it gave out: 834275.

Alternatively, $\sum_{i=1}^{50}\sum_{j=i}^{50}ij=\sum_{i=1}^{50}\l eft\{\sum_{j=1}^{50}ij-\sum_{j=1}^{i-1}ji\right\}$. The inner sum is $i\left(\cdot\sum_{j=1}^{50}j-\sum_{j=1}^{i}j\right)=i\left(\frac{50\cdot 51}{2}-\frac{i\cdot (i+1)}{2}\right)$