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From: Matt J on 25 Mar 2010 16:03
James Allison <james.allison(a)mathworks.com> wrote in message <hogcta$dqh$1(a)fred.mathworks.com>...
> How about:
>
> sum(sum((2:7)'*(1:5).^2))
========

which can be simplified a bit more to

sum(2:7)*sum((1:5).^2)

Also, the sums of consecutve integers and squares of integers have closed form formulas, so this could also be done without any summation at all:

K=7; T=5;

Result=(K*(K+1)/2-1)*T*(T+1)*(2*T+1)/6
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