2

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I can't go to second step from the first. Can you please explain the way?

I will be obliged if you also explain how to go to third step.

Thank you.

Silent
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    What you do is you write sum the more explicitly by filling in the different values of $j$. By doing so you notice that the index $j=1$ arises $n$ times, $j=2$ arises $n-1$ times, etc... To go from 2 to 3, you just 'complete' the sums so that you get $\sum_{k=1}^n a_kb_k$ each time. To do that you need to add to $\sum_{k=2}^n a_kb_k$ the term $a_1b_1$, but, oh good, you have $n$ of those extra (as you have $na_1b_1$). You continue this way to arrive at the solution... – Jan Keersmaekers Sep 24 '13 at 09:03

1 Answers1

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$\sum_{1\leq j\leq k\leq n}(a_kb_k+a_jb_j)=$

denote $a_ib_i=A_i$ and rewrite the question as

$\sum_{1\leq j\leq k\leq n}(A_j+A_k)=$

$=(A_1+A_1)+$

$+(A_1+A_2)+(A_2+A_2)+$

$................................$

$+(A_1+A_n)+(A_2+A_n)+......+(A_n+A_n)=$

$=(nA_1+A_1)+((n-1)A_2+2A_2)+......+(A_n+nA_n)=$

$=(n+1)A_1+(n+1)A_2+...+(n+1)A_n=$

$=(n+1)(A_1+A_2+...+A_n)=$

$=(n+1)\sum_{k=1}^{n}A_k=$

$=(n+1)\sum_{k=1}^{n}a_kb_k$

Adi Dani
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