Answer to Question #154772 in Discrete Mathematics for ABHISHEK GARG

Question

USING MATHEMATICAL INDUCTION P(n)=1+2²+2ⁿ+........+2ⁿ= 2ⁿ⁺¹ - 1

Answer 1

* I think the question is wrong. The statement will be such as $P(n): 1+2^1+2^2+.......+2^n=2^{n+1}-1$

Let $P(n)$ be a statement such that $P(n): 1+2^1+2^2+.......+2^n=2^{n+1}-1$

Then $P(n)$ is true for $n=0$ as $P(0):1=2^{0+1}-1$

Now we show that the statement is true for smallest natural number $n=0$ .

Next we have to show for any $k\geq 0$ , if $P(k)$ holds, then $P(k+1)$ also holds.

Assume that for $n=k$ , $P(k)$ is true. $i.e$ $P(k): 1+2^1+2^2+.......+2^k=2^{k+1}-1$ .......(1)

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

Then $1+2^1+2^2+.......+2^k+2^{k+1}=(2^{k+1}-1)+2^{k+1}$ [using (1)]

$=2.2^{k+1}-1$

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

Therefore the statement $P(k+1)$ holds true.

Hence by the principle of mathematical induction the given statement $P(n)$ is true for all $n\in N$ .