# The value of $\sum ij$, where the summation is over all $i$ and $j$ such that $1 \leq i \leq j \leq 10$

by Astrobleme   Last Updated May 16, 2018 14:20 PM

The value of $\sum ij$, where the summation is over all $i$ and $j$ such that $1 \leq i \leq j \leq 10$, is

1. 1320
2. 2640
3. 3025
4. None of the above

How do I organise the numbers so that I can make this extensible for any arbitrary $n$ and not just 10?

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Hint:

Let $$S=\sum_{1\le i< j\le 10}ij$$ observe that $$(1+2+\ldots+10)^2=1^2+2^2+\ldots+10^2+2S$$ The sum you are looking for is $$\sum_{1\le i\le j\le 10}ij=S+1^2+2^2+\ldots+10^2$$

Ángel Mario Gallegos
February 22, 2017 16:43 PM

HINT Let $S_n = \sum_{i=1}^n i$. Then, $$\sum_{i=1}^n \sum_{j = i}^n ij = \sum_{i=1}^n i \sum_{j = i}^n j = \sum_{i=1}^n i (S_n - S_{i-1}) = S_n \sum_{i=1}^n i - \sum_{i=1}^n i S_{i-1} = S_n^2 - \sum_{i=1}^n i S_{i-1}$$ Can you find $S_n$ and finish the simplification?

gt6989b
February 22, 2017 16:44 PM

You can reorganize the sum as

$$\sum_{1\leq i\leq j\leq n}ij=\sum_{j=1}^n j\sum_{i=1}^j i=\sum_{j=1}^n\frac{j^2(j+1)}{2}=\frac{1}{2}\sum_{j=1}^nj^3+\frac{1}{2}\sum_{j=1}^nj^2=\frac{n^2(n+1)^2}{8}+\frac{n(n+1)(2n+1)}{12}$$

Michael Burr
February 22, 2017 16:44 PM

Σij, where the summation is over all i and j such that 1 ≤ i < j ≤ 10 = 1*(2 + 3 + … + 10) + 2(3 + 4 + … + 10) + 3(4 + 5 + … + 10) + 4(5 + 6 +… + 10) + 5(6 + 7 + … + 10) + 6(7 + 8 + 9 + 10) + 7(8 + 9 + 10) + 8(9 + 10) + 9*10 = (9/2){2*2 + (9 – 1)1} + 2(8/2){2*3 + (8 – 1)*1} + 3(7/2){2*4 + (7 – 1)*1} + 4(6/2){2*5 + (6 – 1)*1} + 5(5/2){2*6 + (5 – 1)*1} + 6*34 + 7*27 + 8*19 + 90 = 54 + 104 + 147 + 180 + 200 + 204 + 189 + 152 + 90 = 1320

Kuntal Sau
May 16, 2018 14:19 PM