Double Summation

[nnugochi]

I'm still new to this, but can someone show and explain me how this is done please?

 

[Otis]

I'm still new to this, but can someone show and explain me how this is done please?

Hi nnugochi. Sure thing. There are many video examples online, explaining double summations in general and properties of double summations (search keywords: double summation properties).

?

An enthusiastic guy demonstrates the usefulness of the Double Sum to Product of Sums property, in the video below.

$\;$

 

[HallsofIvy]

First, if this really is $\displaystyle \sum_{i=1}^j\sum_{j=1}^5 3ij$ then the "inner sum", $\displaystyle \sum_{j= 1}^5 3ij= 3i(1)+ 3i(2)+ 3i(4)+ 3i(5)= 3i(1+ 2+ 3+ 4+ 5)= 3i(15)= 45i$ so that the "outer sum" is $\displaystyle \sum_{i=1}^j 45i= 45\sum_{i= 1}^j i= 45\left(\frac{j(j+1)}{2}\right)$.

However, I suspect that you have the sums reversed and that it was really supposed to be $\displaystyle \sum_{j=1}^5\sum_{i=1}^j 3ij$. That is the same as $\displaystyle \sum_{j=1}^5 3j\left(\sum_{i=1}^j i\right)$.

And again we can use the fact that $\displaystyle \sum_{i=1}^j i= \frac{j(j+1)}{2}$. To see that, write the sum as
$\displaystyle S= 1+ 2+ 3+ \cdot\cdot\cdot+ (j-2)+ (j-1)+ j$
and reverse it as
$\displaystyle S= j+(j-1)+(j-2)+ \cdot\cdot\cdot+ 3+ 2+ 1$.

Now add those two "vertically":
$\displaystyle 2S= (1+ j)+ (2+j-1)+ (3+j-2)+\cdot\cdot\cdot+ (j-2+3)+(j-1+2)+(j+1)$
Each of those parentheses is j+1 and there are j of the so the sum is
$\displaystyle 2S= j(j+1)$.

So we have $\displaystyle \sum_{j=1}^5\sum_{i=1}^j 3ij= 3\sum_{j=1}^5\frac{j^2(j+1)}{2}= 3\left(\frac{1(1+1)}{2}+ \frac{4(2+1)}{2}+ \frac{9(3+1)}{2}+ \frac{16(4+1)}{2}+ \frac{25(5+1)}{2}\right)$
$\displaystyle = 3\left(\frac{2}{2}+ \frac{12}{2}+ \frac{36}{2}+ \frac{80}{2}+ \frac{150}{2}\right)= 3(1+ 6+ 18+ 40+ 75)= 3(140)= 420$.

 

[nnugochi]

Hi nnugochi. Sure thing. There are many video examples online, explaining double summations in general and properties of double summations (search keywords: double summation properties).

?

An enthusiastic guy demonstrates the usefulness of the Double Sum to Product of Sums property, in the video below.

$\;$

thank you so much! this helps me a lot and I understand better

 

[nnugochi]

First, if this really is $\displaystyle \sum_{i=1}^j\sum_{j=1}^5 3ij$ then the "inner sum", $\displaystyle \sum_{j= 1}^5 3ij= 3i(1)+ 3i(2)+ 3i(4)+ 3i(5)= 3i(1+ 2+ 3+ 4+ 5)= 3i(15)= 45i$ so that the "outer sum" is $\displaystyle \sum_{i=1}^j 45i= 45\sum_{i= 1}^j i= 45\left(\frac{j(j+1)}{2}\right)$.

However, I suspect that you have the sums reversed and that it was really supposed to be $\displaystyle \sum_{j=1}^5\sum_{i=1}^j 3ij$. That is the same as $\displaystyle \sum_{j=1}^5 3j\left(\sum_{i=1}^j i\right)$.

And again we can use the fact that $\displaystyle \sum_{i=1}^j i= \frac{j(j+1)}{2}$. To see that, write the sum as
$\displaystyle S= 1+ 2+ 3+ \cdot\cdot\cdot+ (j-2)+ (j-1)+ j$
and reverse it as
$\displaystyle S= j+(j-1)+(j-2)+ \cdot\cdot\cdot+ 3+ 2+ 1$.

Now add those two "vertically":
$\displaystyle 2S= (1+ j)+ (2+j-1)+ (3+j-2)+\cdot\cdot\cdot+ (j-2+3)+(j-1+2)+(j+1)$
Each of those parentheses is j+1 and there are j of the so the sum is
$\displaystyle 2S= j(j+1)$.

So we have $\displaystyle \sum_{j=1}^5\sum_{i=1}^j 3ij= 3\sum_{j=1}^5\frac{j^2(j+1)}{2}= 3\left(\frac{1(1+1)}{2}+ \frac{4(2+1)}{2}+ \frac{9(3+1)}{2}+ \frac{16(4+1)}{2}+ \frac{25(5+1)}{2}\right)$
$\displaystyle = 3\left(\frac{2}{2}+ \frac{12}{2}+ \frac{36}{2}+ \frac{80}{2}+ \frac{150}{2}\right)= 3(1+ 6+ 18+ 40+ 75)= 3(140)= 420$.

thank you!! I am more than happy that I can resolve this question and your final answer made me happy that I got the same one too!!