Question

Show that1+2+4+ ... +2^(n-1) = 2^(n) - 1
This is the arithmetic sequence and stuff chapter btw

Answer 1

i think it not AP it is GP. because 2nd term/1st term=3rd term/2nd term

Answer 2

That could be it...

Answer 3

that is 2/1=4/2

Answer 4

if n=4 then 4th term will be 8 and 8/4=4/2

Answer 5

It is indeed a geometric series.

Let \(S=1+2+4+\cdots+2^{n-1}=2^0+2^1+2^2+\cdots+2^{n-1}\).

Multiply through by \(2\) to yield \(2S=2^1+2^2+2^3+\cdots+2^n\).

Observe their difference and watch the dominoes fall: $$\begin{align*}2S-S&=(2^1+2^2+\cdots+2^{n-1}+2^n)-(2^0+2^1+2^2+\cdots+2^{n-1})\\&=2^n-2^0\\&=2^n-1\end{align*}$$
Note that \(2S-S=S\) and thus we've shown \(S=2^n-1\).

Answer 6

Im kind of confused on why you have to multiply it all by 2 then subtract it by S

Answer 7

@Yellowpanda because all of the \(2^1,2^2,2^3,\dots,2^{n-1}\) terms cancel out.

Answer 8

Hmmm. I guess I understand it now ^_^. Thanks!

Answer 9

This is actually based on the general argument used to determine the sum of any geometric series. Consider the sequence \(\{a_n\}\) where \(a_n=a_1r^{n-1}\). We have for the sum of the first \(n\) terms: $$\begin{align*}S_n&=a_1+a_1r+a_1r^2+\cdots+a_1r^{n-1}\\&=a_1(1+r+r^2+\cdots+r^{n-1})\end{align*}$$ Consider, then, multiplying through by our common ratio \(r\) $$\begin{align*}rS_n&=ra_1(1+r+r^2+\cdots+r^{n-1})\\&=a_1(r+r^2+\cdots+r^{n-1}+r^n)\end{align*}\\$$Observe their difference is:$$\begin{align*}rS_n-S_n&=a_1(r+r^2+\cdots+r^{n-1}+r^n)-a_1(1+r+r^2+\cdots+r^{n-1})\\&=a_1(-1+r-r+r^2-r^2+\cdots+r^{n-1}-r^{n-1}+r^n)\\&=a_1(-1+r^n)=a_1(r^n-1)\end{align*}$$
Note our left-hand side is \(rS_n-S_n=S_n(r-1)\) so we have $$S_n(r-1)=a_1(r^n-1)\\S_n=a_1\frac{r^n-1}{r-1}$$

Answer 10

Then, for the infinite series, we notice for \(|r|<1\) that \(r^n\to\infty\) as \(n\to\infty\), meaning \(S_n=\frac{r^n-1}{r-1}\to S_\infty=\frac{-1}{r-1}=\frac1{1-r}\).

Answer 11

Very detailed! Thank you very much! :)

Answer 12

OOPS, I meant \(r_n\to0\) which should make sense. Repeatedly multiplying by a fraction smaller than \(1\) intuitively gives us less and less each time. As we start summing up more and more terms, these terms at the end then get smaller and smaller, closer and closer to 0.

Answer 13

... which reflects \(1+\frac12+\frac14+\cdots=2\)