Geometric Sum

[Moridin]

Homework Statement

Calculate the following sum:

$$\sum_{k=1}^{10} 3 \cdot 2^{k}$$

The Attempt at a Solution

$$\sum_{k=1}^{10} 3 \cdot 2^{k} = 3 \cdot \sum_{k=1}^{10} 2^{k} = 3 \cdot (2^{10} - 1)$$

Although it seems that I am missing something.

 

[D H]

$$\sum_{k=1}^{10} 3 \cdot 2^{k} = 3 \cdot \sum_{k=1}^{10} 2^{k} = 3 \cdot (2^{10} - 1)$$

Although it seems that I am missing something.

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

 

[Moridin]

Right, if you divide a term in the sum by the previous term, you get 2, not 1? So a factor of 2 is missing.

 

[Gib Z]

Do you know $$x^n - y^n = (x-y)(x^{n-1} + x^{n-1}y + ... + y^{n-1} )$$ ?

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 =]

 

[spideyunlimit]

**Edited by user.

 

[Gib Z]

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

 

[spideyunlimit]

ok sorry buddy!

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

 

[HallsofIvy]

Homework Statement

Calculate the following sum:

$$\sum_{k=1}^{10} 3 \cdot 2^{k}$$

The Attempt at a Solution

$$\sum_{k=1}^{10} 3 \cdot 2^{k} = 3 \cdot \sum_{k=1}^{10} 2^{k} = 3 \cdot (2^{10} - 1)$$

Although it seems that I am missing something.

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
$$\sum_{k=0}^n a 2^k= a (2^{n+1}- 1)$$

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

 

[Moridin]

Let's see.

The general formula is

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

Now the terms in the sum

$$\sum_{k=1}^{10} 3 \cdot 2^{k} = 6 + 12 + 24 + ... + 3072$$

This means that

n = 9 (10-1)
r = 2 (taking the n:th term/(n-1):th term)
a = 6 (the starting term)

So

$$\sum_{k=1}^{10} 3 \cdot 2^{k} = 6\frac{1-2^{9+1}}{1-2} = 6\frac{1-2^{10}}{1- 2} = 6(2^{10} - 1)$$

That sounds about right?

Why is n = 9 here? Is that just the number of terms that have an exponent >= 2?

 

[spideyunlimit]

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.

 

[HallsofIvy]

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>

 

[spideyunlimit]

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