Geometric Series

[chrislam5459]

Can anyone show me how this can be done without actually adding everything term by term?
Thanks!

 

[MarkFL]

I would write:

[MATH]\sum_{n=4}^{14}(2^n)=2^4\sum_{n=0}^{10}(2^n)=2^4(2^{11}-1)=16(2047)=32752[/MATH]

 

[Deleted member 4993]

View attachment 17689
Can anyone show me how this can be done without actually adding everything term by term?
Thanks!

You could use the formula of sum for a geometric series.

 

[chrislam5459]

I would write:

[MATH]\sum_{n=4}^{14}(2^n)=2^4\sum_{n=0}^{10}(2^n)=2^4(2^{11}-1)=16(2047)=32752[/MATH]

Why is it 14 instead of 12?

 

[Deleted member 4993]

Why is it 14 instead of 12?

That is solution of a similar example problem - not exactly your problem.

 

[MarkFL]

Why is it 14 instead of 12?

Sorry, my aging eyes got me again.

 

[pka]

View attachment 17689
Can anyone show me how this can be done without actually adding everything term by term?

There are as many ways to do this as there are doers.
\(S=2^4+2^5+\cdots+2^{11}+2^{12}\\2S=2^5+2^6+\cdots+2^{12}+2^{13}\\\text{thus subtracting}\\S=2^{13}-2^4\) See here
In general: \(M > N,\quad \sum\limits_{n = N}^M {{2^n}} = {2^{M + 1}} - {2^N}\)

But given today's technology, this is really only an interesting artifact of mathematical history. See Here.