I am going through my textbook and have come across this question that asks to work out the sigma, this sigma is troubling me, I am not understanding how to do the "to the power of I + 2" part. Any help would be great, thanks. $$\sum_{i=2}^7\,2^{i+2}$$
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Do you mean: $\sum_{i=2}^7,2^{i+2}$? – lulu Mar 28 '17 at 22:31
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Yes that's correct lulu, i was only wording it the same way as the textbook has. – Richard Motion Mar 28 '17 at 22:32
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Well, what's the problem? It's only six terms ...you could do it by hand. The usual trick would be to multiply it by $2$ and compare the two sums, but really it's not hard to just do it out. – lulu Mar 28 '17 at 22:34
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$2^4 + 2^5 + 2^6 + 2^7 + 2^8 + 2^9 = 16 + 32 + 64 + 128 + 256 + 512 = 1023 - 31 = 992$ – Yunus Syed Mar 28 '17 at 22:35
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@YunusSyed, I think you meant to subtract $15$, not $31$. – Barry Cipra Mar 29 '17 at 00:06
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yes, silly mistake – Yunus Syed Mar 29 '17 at 01:43
3 Answers
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Observe that $$ \sum_{i=2}^72^{i+2}=\sum_{i=2}^72^i2^2=2^2\sum_{i=2}^72^i. $$ What happens is just that the power is $2$ more than the current value of $i$. Initially, $i=2$, so the power starts at $i+2=4$. If you want to further reduce the sum, let $j=i-2$, then $i=j+2$ and $$ 2^2\sum_{i=2}^72^i=2^2\sum_{j=0}^52^{j+2}=2^2\sum_{j=0}^52^j2^2=2^4\sum_{j=0}^52^j. $$ You can calculate this by hand or use the geometric sum formula at this point.
Michael Burr
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Using $2^{x}=(2-1)2^{x}=2^{x+1}-2^{x}$. We are trying to find,
$$2^{2+2}+2^{3+2}+2^{4+2}+2^{5+2}+2^{6+2}+2^{7+2}$$
$$=2^{4}+2^{5}+2^{6}+2^{7}+2^{8}+2^{9}$$
$$=(\color{red}{2^{5}}-2^{4})+(\color{blue}{2^{6}}-\color{red}{2^{5}})+(\color{green}{2^{7}}-\color{blue}{2^{6}})+(\color{orange}{2^{8}}-\color{green}{2^{7}})+(\color{purple}{2^{9}}-\color{orange}{2^{8}})+(2^{10}-\color{purple}{2^{9}})$$
Everything cancels except for,
$$-2^{4}+2^{10}$$
$$=1008$$
Ahmed S. Attaalla
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Um... just do it
$\sum_{i=2}^7 2^{i+2} = 2^{2+2} + 2^{3+2} + 2^{4+2} + .... + 2^{7+2}$
... or were you not supposed to do it directly and figure a formula.
$b^{i+2}= b^ib^2$
So $\sum_{i=2}2^{i+2} = \sum_{i=2}^72^i*4 = 4\sum_{i=2}^7 2^i$ Can you figure out what $\sum_{i=2}^7 2^i$ is?
Did you know that $\sum_{i=0}^k 2^i = 2^{k-1} -1$? If you didn't can you figure out why $1 +2 +4 + 8 + ....... + 2^{k-1} = 2^k - 1$?
Knowing that... $\sum_{i = 2}^7 2^i$ suppose your replace all the $i$ with $j = i-2$ so $i = j+2$. $\sum_{i=2;j = 2-2=0}^{7-2=5} 2^{j+2}=\sum_{j=0}^5 2^22^j$. Can you figure that out?
fleablood
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