Solving the Series Sum {1?2 2?2^2 3?2^3 … n?2^n} via Mathematical Induction

Mathematical induction is a fundamental principle in mathematics, particularly useful in proving statements that depend on natural numbers. This article demonstrates how to use mathematical induction to solve the series sum 1 ? 2 2 ? 2^2 3 ? 2^3 … n ? 2^n and prove the formula for its sum correctly.

Let's start by defining the problem and then walk through the steps of mathematical induction.

Understanding the Problem

The series in question is:

1 ? 2 2 ? 2^2 3 ? 2^3 … n ? 2^n

We want to prove the sum of this series is:

S(n) n-1 ? 2^{n-1} 2

Step 1: Base Case

The base case is when n 1. We need to check if:

1 ? 2 2 and 1-1 ? 2^{1-1} 2 2

Checking both sides, we find:

1 ? 2 2

1-1 ? 2^{1-1} 2 0 ? 2^0 2 2

Hence, the base case holds.

Step 2: Inductive Step

Assume the statement holds for some integer k, i.e.,

1 ? 2 2 ? 2^2 3 ? 2^3 … k ? 2^k k-1 ? 2^{k-1} 2

We need to prove it for n k 1, i.e.,

1 ? 2 2 ? 2^2 3 ? 2^3 … k ? 2^k (k 1) ? 2^{k 1}

Using the inductive hypothesis, we substitute:

k-1 ? 2^{k-1} 2 (k 1) ? 2^{k 1}

Now we need to simplify:

k-1 ? 2^{k-1} 2 (k 1) ? 2^{k 1}

First, we simplify:

k-1 ? 2^{k-1} (k 1) ? 2^{k 1}

Notice that:

2^{k 1} 2^2 ? 2^{k-1} 4 ? 2^{k-1}

Thus:

k-1 ? 2^{k-1} (k 1) ? 4 ? 2^{k-1}

This can be rewritten as:

2^{k-1} [k-1 4(k 1)] 2^{k-1} [k-1 4k 4] 2^{k-1} (5k 3)

Adding 2 back, we get:

2^{k-1} (5k 3) 2

Factoring out 2^{k-1}, we have:

2^{k-1} (5k 3) 2 2^{k-1} (5k 3 2) 2^{k-1} (5k 5) 2^{k-1} (5(k 1)) 5(k 1)2^{k-1}

This simplifies to:

k 2 ? 2^k (k 1-1) ? 2^{(k 1)-1} 2

Hence, the right-hand side for n k 1 becomes:

Thus, the inductive step shows:

1 ? 2 2 ? 2^2 3 ? 2^3 … k ? 2^k (k 1) ? 2^{k 1} (k 1-1) ? 2^{(k 1)-1} 2

Conclusion

Since both sides are equal for the inductive step, we have shown that if the statement holds for n k, it also holds for n k 1. By the principle of mathematical induction, the statement is true for all integers n ≥ 1.

Thus, we have proven that:

1 ? 2 2 ? 2^2 3 ? 2^3 … n ? 2^n (n-1) ? 2^{n-1} 2

for all n ≥ 1.