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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?

3 Answers 3

2

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?

5

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 $$\sum_{1\le i\le j\le 10}ij=S+1^2+2^2+\ldots+10^2$$

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    What happened to the last line? It's a bit confusing. – 2017-02-22
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    In the last line, the inequality $<$ became $\leq$. What happens is that in the product $(1+2+\cdots+10)^2$ you get both $2*3$ and $3*2$. Since $i\leq j$, we'll collect those and divide by $2$. Unfortunately, the squares like $3*3$ appear only once, so we'll need to add new a new set of squares so that we double them up and can divide by $2$. – 2017-02-22
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You can reorganize the sum as

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

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    But the answer given is $1320$ and not $1705$ as your formula provides. :/ – 2017-02-22
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    Nevermind. Your answer is correct. I tested it using a simple Java program. The answer *is* 1705. – 2017-02-22
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    You get 1320 with the condition $1\leq i$i$ is *strictly* less than $j$ rather than $i$ is less than or equal to $j$. – 2017-02-22
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    Oh. Silly mistake on my end. – 2017-02-22