Simplify $\sum_{i=1}^n(\sum_{j=i}^n j)$

Question

$\sum_{i=1}^n(\sum_{j=i}^n j)$

This really is lame, but i couldn't figure out how to work with this one.

I can easily tell that $\sum_{i=1}^ni= \dfrac{n(n+1)}{2}$, and that the summation i am trying to simplify should be something like - $\sum_{i=1}^n(\sum_{j=i}^n j) = \dfrac{n(n+1)}{2} +\dfrac{n(n+1)}{2} -1 +\dfrac{n(n+1)}{2}-(1+2)\,+...+\,n$

Any clever ways to simplify this expression ? Thank you!

Answer 1

Let $$ S=\sum_{i=1}^n\sum_{j=1}^nj=n\frac{n(n+1)}{2}=\frac{n^2(n+1)}{2}. $$ Next, consider $$ T=\sum_{i=1}^n\sum_{j=1}^{i-1}j=\sum_{i=1}^n\frac{i(i-1)}{2}=\frac{1}{2}\sum_{i=1}^ni^2-\frac{1}{2}\sum_{i=1}^ni\\ =\frac{n(n+1)(2n+1)}{12}-\frac{n(n+1)}{4}. $$ Then, it remains to compute $S-T$. Do you notice what the result equals to?

Answer 2

HINT

Note that

$$\sum_{j=i}^n j=\sum_{j=1}^n j-\sum_{j=1}^{i-1} j$$

Answer 3

You can interchange the two sums. This is a very powerful technique for simplifying double summations. To do so, notice that the sum is over all pairs $(i,j)$ with $1 \le i \le j \le n$, so we can say that

$$ \sum_{i = 1}^n \sum_{j = i}^n j = \sum_{j = 1}^n \sum_{i = 1}^j j = \sum_{j = 1}^n j^2 = \frac{n(n+1)(2n+1)}{6}. $$

Answer 4

We have

\begin{align*} \sum_{i=1}^n\sum_{j=i}^n j&=\sum_{i=1}^n\frac{(n-i+1)(i+n)}{2}\\ &=\frac{n(n+1)}{2}\sum_{i=1}^n1+\frac{1}{2}\sum_{i=1}^ni-\frac{1}{2}\sum_{i=1}^ni^2\\ &=\frac{n(n+1)}{2}(n)+\frac{1}{2}\left[\frac{1}{2}n(n+1)\right]-\frac{1}{2}\left[\frac{1}{6}n(n+1)(2n+1)\right]\\ &=\frac{1}{12}n(n+1)\left[6n+3-(2n+1)\right]\\ &=\frac{1}{6}n(n+1)(2n+1) \end{align*}

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Alternatively, note that $\displaystyle \sum_{i=1}^n\sum_{j=i}^n a_{ij}=\sum_{j=1}^n\sum_{i=1}^j a_{ij}$.

\begin{align*} \sum_{i=1}^n\sum_{j=i}^n j&=\sum_{j=1}^n\sum_{i=1}^j j\\ &=\sum_{j=1}^n j^2\\ &=\frac{1}{6}n(n+1)(2n+1) \end{align*}

It is easy to see that the sum is $1^2+2^2+3^2+\cdots+n^2$.