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).I'm still new to this, but can someone show and explain me how this is done please?
thank you so much! this helps me a lot and I understand betterHi nnugochi. Sure thing. There are many video examples online, explaining double summations in general and properties of double summations (search keywords: double summation properties).
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An enthusiastic guy demonstrates the usefulness of the Double Sum to Product of Sums property, in the video below.
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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!!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 \(\displaystyle \sum_{j=1}^5\sum_{i=1}^j 3ij= 3\sum_{j=1}^5\frac{j^2(j+1)}{2}= 3\left(\frac{1(1+1)}{2}+ \frac{4(2+1)}{2}+ \frac{9(3+1)}{2}+ \frac{16(4+1)}{2}+ \frac{25(5+1)}{2}\right)\)
\(\displaystyle = 3\left(\frac{2}{2}+ \frac{12}{2}+ \frac{36}{2}+ \frac{80}{2}+ \frac{150}{2}\right)= 3(1+ 6+ 18+ 40+ 75)= 3(140)= 420\).