Hi Zamira,
I want to state the problem more precisely.
Prove that 1+2+2^{2}+2^{3}+...+2^{n1} = 2^{n}  1 for n = 1, 2, 3, ...
There are two steps in a proof by induction, first you need to show that the result is true for the smallest value on n, in this case n = 1.
When n = 1 the left side has only one term, 2^{n1} = 2^{11} = 2^{0} = 1. The right side is 2^{n}  1= 2^{1}  1 = 1. Thus the statement is true for n = 1.
The second step is the inductive step. You need to show that if the statement is true for any particular value of n it is also true for the next value of n. If you can do this then, since it is true for n = 1 it will also be true for the next value of n, n = 2. Then, since it is true for n = 2 it will also be true for n = 3, and so on. To show the inductive step I am going to assume that the statement is true for the particular value of n, n = k. Here is my proof.
Suppose that the statement is true for n = k, that is
1+2+2^{2}+2^{3}+...+2^{k1} = 2^{k}  1 equation 1
I need to prove that the statement is true for n = k + 1, that is I need to prove that
1+2+2^{2}+2^{3}+...+2^{(k+1)1} = 2^{k+1}  1 equation 2
The left side of equation 2 is
1+2+2^{2}+2^{3}+...+2^{(k+1)1} = 1+2+2^{2}+2^{3}+...+2^{k}
The next to the last term in this sum is 2^{k1} and hence I can write the left side of equation 2 as
1+2+2^{2}+2^{3}+...+2^{k1} + 2^{k}
But I know that 1+2+2^{2}+2^{3}+...+2^{k1} = 2^{k}  1 (this is equation 1) so the left side of equation 2 is is
1+2+2^{2}+2^{3}+...+2^{k1} + 2^{k} = 2^{k}  1 + 2^{k} = 2(2^{k})  1 = 2^{k+1} 1
Hence we have shown that the left side and right side of equation 2 are equal.
This completes the inductive step and hence, by induction, I have proved that
1+2+2^{2}+2^{3}+...+2^{n1} = 2^{n}  1 for n = 1, 2, 3, ...
Penny
