#Can someone help
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Suppose i = 1. How many terms will the inner sum have?
if "i = 1 " there will be n(n+1)/2 additons
Oh.
Well, rather, for i = 1 there's just one term, but for all values of i we get 1 + 2 + ... + n = n(n + 1)/2 terms.
So if I'm not mistaking, there will be n(n+1)/2 multiplications and (n+2)(n-1)/2 additions.
Ohh, sorry. I didn't notice that we had a(i)b(j), I thought we just had a(i, j).
No worries!
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