Evaluation a non-linear series?

**chancey** · September 5th 2006, 05:06 PM

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

Thanks

**ThePerfectHacker** · September 5th 2006, 05:47 PM

Originally Posted by chancey

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

Thanks

This is a geometric series.
---
$(2^1+2^2+...+2^{20})-(1+1...+1)$
Thus,
$2(1+2+2^2+...+2^{19})-20$
Use geometric sum formula,
$2(2^{20}-1)-20$
Simplify,
$2^{21}-22$

**chancey** · September 5th 2006, 06:50 PM

Originally Posted by ThePerfectHacker

This is a geometric series.
---
$(2^1+2^2+...+2^{20})-(1+1...+1)$
Thus,
$2(1+2+2^2+...+2^{19})-20$
Use geometric sum formula,
$2(2^{20}-1)-20$
Simplify,
$2^{21}-22$

Right, but thats not the answer in the back of the book...

I did it this way:
$\sum_{n=1}^{20} (2^n - 1) = \sum_{n=1}^{20} (2^n) - 20$

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

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

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

Which is the same answer as yours, however the answer in the book says $2097170$ . They must of added the 20 instead of subtracted it. OK thanks

**ThePerfectHacker** · September 5th 2006, 06:53 PM

Originally Posted by chancey

Right, but thats not the answer in the back of the book...

I did it this way:
$\sum_{n=1}^{20} (2^n - 1) = \sum_{n=1}^{20} (2^n) - 20$

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

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

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

Which is the same answer as yours, however the answer in the book says $2097170$ . They must of added the 20 instead of subtracted it. OK thanks

Then the book is inncorect.

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