[size=150]If n ∈ N, then 2[sup]1[/sup]+2[sup]2[/sup] +2[sup]3[/sup] +··· +2[sup]n[/sup] = 2[sup]n+1[/sup] −2. [br][br]Proof. The proof is by mathematical induction.[br][br](1) When n = 1, this statement is 2[sup]1[/sup] = 2[sup]1+1[/sup] −2, or 2 = 4−2, which is true.[br][br](2) Now assume the statement is true for some integer n = k ≥ 1, that is assume [br][br]2[sup]2[/sup] + 2[sup]2[/sup] + 2[sup]3[/sup] + ··· + 2[sup]k[/sup]= 2[sup]k+1[/sup] − 2. Observe this implies that the statement is true for n = k +1, as[br]follows:[br][br]2[sup]1[/sup] +2[sup]2[/sup] +2[sup]3[/sup] +··· +2[sup]k[/sup]+2[sup]k+1[/sup] =[br][br](2[sup]1[/sup] +2[sup]2[/sup]+2[sup]3[/sup] +··· +2[sup]k[/sup] )+2[sup]k+1[/sup] = [br]                    [br]2[sup]k+1[/sup] −2+2[sup]k+1[/sup] = 2·2[sup]k+1[/sup] −2[br][br] = 2[sup]k+2[/sup]−2 [br][br]                       = 2[sup](k+1)+1[/sup]−2.[br][br]Thus, we have 2[sup]1[/sup]+2[sup]2[/sup]+2[sup]3[/sup]+···+2[sup]k[/sup]+2[sup]k+1[/sup] = 2[sup](k+1)+1[/sup] −2, so the statement is true for n = k+1.[br][br] Thus, the result follows by mathematical induction[/size]

[size=150]The proof uses mathematical induction to show that the[br]formula 2[sup]1[/sup] + 2[sup]2[/sup] + 2[sup]3[/sup] + ... + 2n = 2[sup]n+1[/sup]− 2 holds for every positive integer n.[br][br]Step 1: Base Case (n = 1)[br]The proof begins by verifying the formula for the base case,[br]where n = 1. In this case, the left-hand side of the equation is 2[sup]1[/sup]= 2, and the right-hand side is[br] 2[sup]1+1[/sup] − 2 = 2. Since both sides are equal, the formula is true for n = 1.[br][br]Step 2: Inductive Hypothesis[br]The proof then assumes that the formula is true for some[br]integer n = k ≥ 1. That is, it assumes that 22 + 22 + 23 + ... + 2k = 2[sup]k+1[/sup]− 2 is true.[br][br]Step 3: Inductive Step (n = k + 1)[br]The proof aims to show that if the formula is true for n =[br]k, then it must also be true for n = k + 1. To do this, the proof considers the[br]sum 2[sup]1[/sup] + 2[sup]2[/sup] + ... + 2k + 2[sup]k+1[/sup].[br][br]Using the inductive hypothesis, the sum 2[sup]1[/sup] + 2[sup]2[/sup]+ ... + 2k can be replaced with 2[sup]k+1[/sup] − 2. Thus, the entire sum becomes 2[sup]k+1[/sup] − 2 + 2[sup]k+1[/sup].[br][br]Next, the proof simplifies the expression by combining like terms. The sum becomes 2[sup](2k+1)[/sup] − 2, which is equal to 2[sup]k+2[/sup]− 2.[br][br]Finally, using the exponent rule that 2[sup](2k+1)[/sup]is equal to 2[sup]k+2[/sup], the expression becomes 2[sup]k+1[/sup]+1 − 2, which is the right-hand side of the formula for n = k + 1.[br][br]Since the inductive step shows that if the formula is true for n = k, then it must also be true for n = k + 1, the proof establishes that the formula 2[sup]1[/sup] + 2[sup]2[/sup] + 2[sup]3[/sup] + ... + 2n = 2[sup]n+1[/sup]− 2 holds for every positive integer n.[/size]