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 some `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^(k+1)-1 ` as required.
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:
I guess this answer will help you to understand easily .....please check my answer in the image file :)