Answer
To demonstrate (prove) that $H_n$ = $2^n-1$ ,
where $H_n$ is defined recursively by the relation $H_n$ = $2\times H_{n-1}+1$ and denotes the minimum number of moves required to solve a tower of hanoi problem with n pegs.
Work Step by Step
Method: Proof by induction
For n = 1 ,
$2^1$ - 1 = 1
For a $k\geq2$, let us assume that $H_k$ = $2^k-1$.
We have to prove that $H_{k+1}$ = $2^{k+1}-1$ .
Given that $H_n$ = $2\times H_{n-1}+1$,
⇒ $H_{k+1}$ = $2\times H_{(k+1)-1}+1$
⇒ $H_{k+1}$ = $2\times H_{k}+1$
⇒ $H_{k+1}$ = $2\times (2^k-1) + 1$
⇒ $H_{k+1}$ = $ 2^{k+1}-2 + 1$
Thus,
$H_{k+1}$ = $ 2^{k+1}-1$ which was to be proved.
Hence, by the domino effect of the induction, the formula derived
in Example 2 has been proven
