**1 + 2 + 22 + … + 2n = 2n +1 – 1 for all natural numbers .**

###### Answers:

Let P(n): 1 + 2 + 2^{2} + … + 2n = 2n +1 – 1, for all natural numbers n

P(1): 1 =2^{0} + 1 – 1 = 2 – 1 = 1, which is true.

Hence, ,P(1) is true.

Let us assume that P(n) is true for some natural number n = k.

P(k): l+2 + 2^{2}+…+2^{k} = 2^{k+1}-l (i)

Now, we have to prove that P(k + 1) is true.

P(k+1): 1+2 + 2^{2}+ …+2k + 2k+1

= 2^{k +1} – 1 + 2^{k+1 } [Using (i)]

= 2.2^{k+l}– 1 = 1

= 2^{(k+1)+1}-1

Hence, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for any natural number n.