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I am trying to figure out how to write the summation for 2+4+8+16+...+1024. Clearly my answer below is wrong by 1:

Attempted answer

I tried to apply the following summation formula but I guess I did not do it properly:

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Please keep in mind I am not a mathy person. This is for an algorithm analysis course.

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    Note that your sum starts at $i=1$ but you try to use the formula for a sum starting at $i=0$. – 2017-01-28
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    The first sum is from $i=1 \to 10$, the second sum is from $i=0 \to 10$. Thus, you are missing out by the number $1$, which is just $2^0$. – 2017-01-28
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    I noticed the difference in i but I am not really sure how to be able to change the summation formula to account for that, like how I can change the summation formula to consider different values of i :S – 2017-01-28
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    @user3451821 $\sum_{i=0}^n a_n = a_0 + \sum_{i=1}^n a_n$. – 2017-01-28
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    Just to clarify, if my i was in fact 2. Would I have to add a0, a1 to the summation in that case? – 2017-01-28
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    Yes, you would have to add the extra terms $a_0,a_1$ in that case. – 2017-01-28

5 Answers 5

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Your summation is $$ \sum_{i=1}^{10}2^i= 2\sum_{i=1}^{10}2^{i-1}= 2\sum_{i=0}^{9}2^{i}= 2(2^{10}-1)= 2^{11}-2=2046 $$ Note the difference in the starting point between your problem and the formula you quote.

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\begin{align} 2+4+8+\ldots+1024 &= x \\ 2+2+4+8+\ldots+1024 &= 2+ x \\ (2+2)+4+8+\ldots+1024 &= 2+x \\ 4+4+8+\ldots+1024 &= 2+x \\ (4+4)+8+\ldots+1024 &= 2+x \\ 8+8+\ldots+1024 &= 2+x \\ &\ \vdots\\ 1024+1024 &= 2+x \\ 2048 &= 2+x \\ -2+2048 &= x \end{align}

I hope this helps $\ddot\smile$

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    This is good method. – 2017-01-28
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The general formula for the sum of consecutive terms of a geometric series, not necessarily starting at $1$ is the following: $$\sum_{k=m}^{k=n}q^k=\frac{q^{n+1}-q^m}{q-1}.$$ Factoring $q^m$ in the sum leads to applying the base formula with last exponent $n-m$, but it's worth knowing it for its own sake.

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"I noticed the difference in i but I am not really sure how to be able to change the summation formula to account for that, like how I can change the summation formula to consider different values of i :S "

The sumation index makes all the difference.

You can always do this:

$\sum_{i=0}^n a_i = \sum_{i=0}^k a_i + \sum_{i=k+1}^n a_i$ for any $0 < k < n$

So $\sum_{i=0}^2 2^i = 2^0 + \sum_{i=1}^n 2^i=$

$ 1 + \sum_{i=1}^n 2^i$ so $\sum_{i=1}^n 2^i = \sum_{i=0}^2 2^i - 1 = $

$(2^{n+1}-1) - 1 = 2^{n+1} - 2$.

....

and if you are really clever you can to this.

$\sum_{i=1}^n 2^i$; for each $i$ let's let $i = k+ 1$ so....

$\sum_{i=1;\text{ or } k = 0}^{i = n; \text{ or } k=n-1} (2^i=2^{k+1}) =$

$\sum_{k = 0}^{n-1} 2^{k+1} =$

$\sum_{k=0}^{n-1} 2*2^k = 2\sum_{k=0}^{n-1}2^k=$

$2(2^n - 1) = 2^{n+1} - 2$.

You can sleep on the second part.

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\begin{align} 2+4+8+\ldots+1024 &= \\ (-1+1)+2+4+8+\ldots+1024 &= \\ -1+(1+2+4+8+\ldots+1024) &= \\ -1 + \sum_{i=0}^{10}2^i \\ \end{align}

Now you can use the formula you mentioned.