# 1+2+4+...+2^(n-1)=2^(n)-1

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Show that `1+2+4+ * * * + 2^(n-1)=2^(n)-1 `

We will proceed using mathematical induction:

(1) Base case: if n=1 then `2^(1-1)=2^0=1=2^1-1 `

(2) Inductive hypothesis: Assume that for someĀ `k >= 1 ` the following is true:

`1+2+4+ * * * +2^(k-2)+2^(k-1)=2^k-1 `

(3) We want to show that for such a k, the following is true:

`1+2+4+ * * * + 2^(k-1)+2^(k)=2^(k+1)-1 `

i.e. we want to show that it is true for k+1.

We begin with the left side:

`1+2+4+ * * * +2^(k-1)+2^k `

`=2^k-1+2^k ` by the inductive hypothesis

`=2(2^k)-1 `

`=2^(k+1)-1 ` as required.

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please tell anoter easy method

It helps when you ask a question to put it in context. For instance, for this problem if you are studying or reviewing a unit on sequences and series you might indicate this in your question. Or you could indicate the class level -- often the approach can be discerned from the level of question.

Here, the left side is a geometric series with a(1)=1 and r=2. The formula for the sum of a finite geometric series is `S_n=a_1((1-r^n)/(1-r)) `

We also know the nth term is given by `a_n=a_1(r)^(n-1) `

So the left side is the sum of the first n terms of the geometric series which is equal to:

`S_n=(1)((1-2^n)/(1-2))=2^n-1 `

I guess this answer will help you to understand easily .....please check my answer in the image file :)