Find the sum of the geometric sequence

Question

$$1+ \frac12 + \frac14+\dots+\frac1{2^n}$$

To find the sum of the equation you have to find $n$, the number of terms in the geometric sequence and I don't know how...

The answer in the text book is $2-\frac1{2^n}$.

Answer 1

Clearly, your sum will be $2$ minus a small square or rectangle in the corner.

Answer 2

Starting with $$s_n=\sum\limits_{k=0}^n\dfrac 1{2^k}=1+\frac 12+\frac 14+\frac 18+\cdots+\frac 1{2^n}$$ Multiplying both sides by $2$ gives$$2s_n=2+1+\frac 12+\cdots+\frac1{2^{n-1}}$$ Subtracting, we have$$s_n=2-\frac 1{2^n}$$

Note that this trick can be used on most simple geometric sequences.

Answer 3

Let $S = 1+ \frac12 + \frac14+\dots+\frac1{2^n}$. Then $2S = 2 + 1 + \frac12 + \frac14+\dots+\frac1{2^{n - 1}}$. Subtract the second equation from the first one to get $$S = 2 - \frac1{2^n}.$$

Answer 4

You don't need to find $n$, the question is requesting a formula that lets somebody plug in any arbitrary $n$ and tells her what that sum is.

One way to find that formula, without just looking at patterns and without anything fancy like an induction proof, is to subtract $1$ from the sum and manipulate as follows: $$ S = \sum_{k=0}^n 2^{-k} \\ S-1 = \sum_{k=1}^n 2^{-k} \\ S-1 = \sum_{(k-1) = 0}^{n-1} 2^{-((k-1)+1)} \\ S-1 = \sum_{m = 0}^{n-1} 2^{-(m+1)} \\ S-1 = \frac12 \sum_{m = 0}^{n-1} 2^{-m} \\ S-1 = \frac12 (S - 2^{-n})\\ \frac12 S - 1 = -\frac12 2^{-n}\\ S-2 = -2^{-n} \\ S = 2 - 2^{-n} $$

Answer 5

Write $S_n(x)=\sum_{i=0}^nx^i$. Then $S_n(x)-xS_n(x)=1-x^{n+1}$. Hence $S_n(x)=\frac{1-x^{n+1}}{1-x}$. Thus $S_n(\frac{1}{2})=2-\frac{1}{2^{n}}$.

Answer 6

Results from a well-known high-school formula: $$1-x^{n}=(1-x)(1+x+x^2+\dots x^{n-1})$$ which can be read as well: $$1+x+x^2+\dots x^{n-1}=\frac{1-x^{n}}{1-x}=\frac1{1-x}-\frac{x^{n}}{1-x}.$$

Answer 7

Hint:

$$2=1+1=1+\frac12+\frac12=1+\frac12+\frac14+\frac14=1+\frac12+\frac14+\frac18+\frac18=\\ 1+\frac12+\frac14+\frac18+\cdots\frac1{2^{n-1}}+\frac1{2^n}+\frac1{2^n}.$$