Geometric Sum

Very close but no cigar. How can [itex]\sum_{k=1}^{10} 2^{k} = 2^{10} - 1[/itex]? You are missing something from the right-hand side.

 

Do you know [tex]x^n - y^n = (x-y)(x^{n-1} + x^{n-1}y + ... + y^{n-1} )[/tex] ?

For (x,y) = (2,1) that forms into what you need. Look at the expansion of what you have, 2^10 - 1, and see how that falls different to the sum you want on the RHS, then change it accordingly =]

 

**Edited by user.

 

Forum regulations prohibit giving out answers to homework threads - please just give hints. Edit your post.

 

ok sorry buddy!

@author of thread... look up summation of a GP

 

The formula you are trying to remember is
[tex]\sum_{k=0}^n a r^k= a\frac{1-r^{n+1}}{1- r}[/itex]
For r= 2, so that 1-r= 1-2= -1, that becomes
[tex]\sum_{k=0}^n a 2^k= a (2^{n+1}- 1)[/tex]

There are two important differences between that and the formula you are using.

 

Someone above told me not to give out the answer.. and btw your answer is correct.. and yes n = 10 not 9 . There are 10 terms of the GP.

 

spdeyunlimit, n= 9, not 10 because because the sum starts with k= 1. He is thinking of this as 6+ 6(2)+ 6(2²)+ 6(2<sup>n) with n running from 0 to 9.

Another way to do the problem would be to use the same formula with n= 10, a= 3, but then subtract off 3- to allow for the missing 3(20) term.</sup>

 

for the GP
a, ar, ar^2, ar^3 ... and so on
Sum of first n terms = a(1 - r^n) / (1-r)
where in the first term the power of r is 0.. whereas here it is one.. so yeah i got that.. it will be for n = 9