Recursive function

We claim that $(m+k) \Delta m = 2^k$ for all $k$ and prove it by induction as follows:

1) When $k = 1$ ,

$\begin{aligned} \quad \quad (m+1) \Delta m & = m \Delta m + (m+1) \Delta (m+1) \\ & = 1 + 1 = 2^1 \end{aligned}$ .

$\quad$ Therefore, the claim is true for $k = 1$ .

2) Assuming the claim is true for $k$ , then

$\begin{aligned} \quad \quad (m+k+1) \Delta m & = (m+k) \Delta m + (m+k+1) \Delta (m+1) \\ & = 2^k + 2^k = 2^{k+1} \end{aligned}$ .

$\quad$ The claim is also true for $k + 1$ . Therefore, the claim is true for all $k$ .

$$

Hence, $32 \Delta 11 = (11+21) \Delta 11 = 2^{21} = \boxed{2097152}$