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I have this summation that I need to solve:

$$\sum_{i=2}^{n}\sum_{j=i}^{n} 3i$$

Can someone please help? I have no idea how to start this

gt6989b
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user850676
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1 Answers1

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Just write it out explicitly to see what happens:

\begin{align} \sum_{i=2}^{n}\sum_{j=i}^{n} 3i &=\sum_{i=2}^{n} \underbrace{(3i+3i+\cdots +3i)}_{n-i+1 \textrm{ terms}}\\ &=\sum_{i=2}^n (n-i+1)(3i)\\ &=\sum_{i=2}^n 3n \cdot i - 3i^2+3i\\ &=\bigg((3n+3)\sum_{i=2}^ni\bigg)-3\bigg(\sum_{i=2}^ni^2\bigg) \end{align}

Now work on the following: $$ \sum_{i=2}^n i,\quad \sum_{i=2}^ni^2 $$