How do I evaluate $\displaystyle \sum_{n=1}^{20} (2^n - 1)$ without havng to add up each element?
Thanks
Right, but thats not the answer in the back of the book...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