How do I evaluate $\displaystyle \sum_{n=1}^{20} (2^n - 1)$ without havng to add up each element?

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- Sep 5th 2006, 05:06 PMchanceyEvaluation a non-linear series?
How do I evaluate $\displaystyle \sum_{n=1}^{20} (2^n - 1)$ without havng to add up each element?

Thanks - Sep 5th 2006, 05:47 PMThePerfectHackerQuote:

Originally Posted by**chancey**

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$\displaystyle (2^1+2^2+...+2^{20})-(1+1...+1)$

Thus,

$\displaystyle 2(1+2+2^2+...+2^{19})-20$

Use geometric sum formula,

$\displaystyle 2(2^{20}-1)-20$

Simplify,

$\displaystyle 2^{21}-22$ - Sep 5th 2006, 06:50 PMchanceyQuote:

Originally Posted by**ThePerfectHacker**

I did it this way:

$\displaystyle \sum_{n=1}^{20} (2^n - 1) = \sum_{n=1}^{20} (2^n) - 20$

$\displaystyle \sum_{n=1}^{20} (2^n) = \frac{1-2^{21}}{1-2} - 1 = 2097150$

$\displaystyle \because \sum_{n=0}^{k} (ar^n) = \frac{a(1-r^{k+1})}{1-r}$

$\displaystyle \therefore \sum_{n=1}^{20} (2^n) - 20 = 2097150 - 20 = 2097130$

Which is the same answer as yours, however the answer in the book says $\displaystyle 2097170$. They must of added the 20 instead of subtracted it. OK thanks - Sep 5th 2006, 06:53 PMThePerfectHackerQuote:

Originally Posted by**chancey**