Prove that`1+2+2^(2)+ . . .+2^(n)=2^(n+1)-1`, for all natural number n

1.	Prove that`1+2+2^(2)+ . . .+2^(n)=2^(n+1)-1`, for all natural number n.
Answer» Consider the given statement `P(n):1+2+2^(2)+ . . .2^(n)=2^(n+1)-1`, for natural numbers n. Step I We observe that P(0) is true. `P(1):1=2^(0+1)-1` `1=2^(1)-1` 1=2-1 1=1, which is true. Step II Now, assume that P(n) is true for n=k. So, P(k) : `1+2+2^(2)+ . . .2^(k)=2^(k+1)-1` is true. Step III Now, to prove P(k+1) is true. `P(k+1):1+2+2^(2)+. . .+2^(k)+2^(k+1)` `=2^(k+1)-1+2^(k+1)` `=2*2^(k+1)-1` `=2^(k+1)+1-1` So, P(k+1) is true, whenever P(k) is true. Hence, P(n) is true.

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