Well, let's see. This sum's terms increase by 2 each for each subsequent term, so our summand should be something like 2k where k is the summing index. If we just had 2k, with limits of say 11 to 1010, we'd get

10 \(\displaystyle∑{2}{k}={2}{\left({1}\right)}+{2}{\left({2}\right)}+{2}{\left({3}\right)}+⋯+{2}{\left({10}\right)}={2}+{4}+{6}+⋯+{20}\) k=1

OK. That gets us the going up by two part, but not the fact that it should start at 8. It starts at 2 so what we really need is to add 6 to every term. That should shift up the starting point while keeping the spacing between numbers.

\(10\sum[2k+6]=[2(1)+6]+[2(2)+6]+[2(3)+6]+⋯+[2(10)+6] k=1 =8+10+12+⋯+26\)

Now the only thing left is make sure that the sum stops in the right place. It looks like we are 8 shy. So because we're skipping every other number, that should mean we need 4 more terms. So let's try stopping at the index\( k=14k=14\).

\(14\sum[2k+6]=[2(1)+6]+[2(2)+6]+[2(3)+6]+⋯+[2(14)+6] k=1 =8+10+12+⋯+34\)

And there we go! We're all done.