<s>[INST] Mathematical Induction - Is this correct? The question is : 
Prove that 
$$3 + 6 + 12 + 24 + ... + 3 \cdot 2^{n-1} = 3 \cdot 2^n - 3$$
for all integers $n ≥ 1$
My Solution (I'll skip the basis step and go straight to the inductive step):
II.
Assume that 
$$3 + 6 + 12 + 24 + ... + 3 \cdot 2^{k-1} = 3\cdot 2^k - 3$$
Consider $n=k+1$:
$$3 + 6 + 12 + 24 + ... + 3 \cdot 2^{k-1} + 3 \cdot 2^{(k+1)-1}= 3\cdot 2^{k+1} - 3$$
$$3 + 6 + 12 + 24 + ... + 3 \cdot 2^{k-1} + 3 \cdot 2^k= 3\cdot 2^{k+1} - 3$$
$$3\cdot 2^k - 3 + 3\cdot 2^k = 3\cdot 2^{k+1} - 3$$
Cancelling out the $3$, I get:
$$3\cdot 2^k + 3\cdot 2^k = 3\cdot 2^{k+1}$$
Dividing both sides by $3$, I get:
$$2^k + 2^k = 2^{k+1}$$
$$2^{k+k}  = 2^{k+1}$$
$$2^{2k} ≠ 2^{k+1}$$
III. Therefore, $P(n)$ is not true for all integers $n ≥ 1$
Is my solution correct? I find it weird that we were given a problem that isn't true but I guess that's intentional (unless I'm the one who's wrong.)
 [/INST] This question has been answered by this user (and others) via the comment. In particular quoting the comment:

$2^k+2^k=2×2^k=2^{k+1}$. This is the part where you made mistake. Otherwise it's fine, and you have actually proved it.

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