## Pologirl19 What is the sum of the geometric series one year ago one year ago

1. Pologirl19

$\sum_{x=1}^{10}6(2)^x$

2. satellite73

$6\sum_{x=1}^{10}2^x$ is a start

3. satellite73

$\sum_{x=1}^{10}2^x=2^{11}-2$

4. Pologirl19

where did you get 2^11 - 2

5. Pologirl19

@satellite73

6. General_Gustav

The sum of this series involves factoring out the 6 into the front of the equation, like satellite73 said. I'm sure there is a shortcut to this, but i would write out each $2^{x}$ term with n beginning with 1 and ending with 11 and then add them all together. Probably not the answer that you're looking for, but at least its a start.

7. tkhunny

$$\sum\limits_{x=1}^{10} 2^{x} = 2 + 2^{2} + 2^{3} + ... + 2^{10}$$ Using the usual technique of creating a Sum and multiplying by the common factor, we get $$\dfrac{2 - 2^{11}}{1 - 2} = 2^{11} - 1$$

8. tkhunny

* - 2 Typo, sorry.

9. General_Gustav

^This is the shortcut I was mentioning. :D

10. satellite73

in general $\sum_{k=1}^n r^k = \frac{r^{n+1}-r}{r-1}$

11. satellite73

in your case $$r=2$$ so you get $\frac{2^{11}-2}{2-1}=2^{11}-2$

12. Pologirl19

That is none of my answers the equation for it is $Sn=a1(\frac{ 1- r^n }{ 1 - r })$

13. Pologirl19

@satellite73

14. tkhunny

@Pologirl19 We ran into this problem the other day. 1) The ability to create the formula is worth far more than a list of formulas you don't understand. 2) If you had the formula all along, why didn't you use it? 3) A little flexibility and thought will produce n = 10, a1 = 2, and r = 2 and this is an equivalent expression. Please brush up on your algebra.