Problem 11 - Algebra and Trigonometry

Question

Use mathematical induction to prove that the formula is true for all natural numbers n. $$1 \cdot 2+2 \cdot 2^{2}+3 \cdot 2^{3}+4 \cdot 2^{4}+\cdots+n \cdot 2^{n} =2\left[1+(n-1) 2^{n}\right]$$

Step-by-Step Solution

Answer

The formula is true for all natural numbers \( n \).

Step 1: Base Case

First, we verify that the formula holds true for the smallest natural number, which is 1.Substitute \( n = 1 \) in the given formula:\[ 1 \times 2 = 2 \]According to the right-hand side of the formula, when \( n = 1 \), it becomes:\[ 2[1 + (1 - 1)2^1] = 2[1 + 0] = 2 \]Since both sides are equal, the base case holds true.

Step 2: Inductive Hypothesis

Assume the formula is true for some natural number \( k \). That is, assume:\[ 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \cdots + k \cdot 2^k = 2[1 + (k-1)2^k] \]This assumption is called the inductive hypothesis.

Step 3: Inductive Step

We need to prove the induction step: if the formula holds for \( n = k \), then it must also hold for \( n = k + 1 \).Consider the left-hand side of the formula with \( n = k + 1 \):\[ 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \cdots + k \cdot 2^k + (k+1) \cdot 2^{k+1} \]Substitute the inductive hypothesis:\[ = 2[1 + (k-1)2^k] + (k+1) \cdot 2^{k+1} \]Simplify:\[ = 2 + 2k \cdot 2^k + (k+1) \cdot 2^{k+1} \]Combine the terms:\[ = 2 + 2k \cdot 2^k + k \cdot 2^{k+1} + 2^{k+1} \]\[ = 2 + k \cdot 2^{k+1} + 2^{k+1} \]\[ = 2 + (k+1) \cdot 2^{k+1} \]which can be rewritten as:\[ 2[1 + (k)2^{k+1}] \]Hence, the formula is also true for \( n = k + 1 \).

Step 4: Conclusion

Since the formula is true for the base case \( n = 1 \) and assuming it is true for \( n = k \) implies it is true for \( n = k + 1 \), by the principle of mathematical induction, the formula is true for all natural numbers \( n \).

Key Concepts

Base Case

In mathematical induction, the base case is the starting point. It ensures the formula we want to prove is valid for the smallest value of the variable, typically 1. Here, we begin by checking if the given formula holds true for \( n = 1 \). We substitute \( n = 1 \) into the formula:\[ 1 \times 2 = 2 \]Upon evaluating the right-hand side of the formula with \( n = 1 \), we get:\[ 2[1 + (1 - 1)2^1] = 2[1 + 0] = 2 \]Both evaluations equal 2, confirming the base case.

• The base case shows the formula is true for the starting point, \( n = 1 \).
• This step ensures we have a solid foundation before progressing.

Inductive Hypothesis

The inductive hypothesis is a crucial step, akin to planting the seeds of logic for our arguments. It involves assuming the statement is true for an arbitrary natural number, say \( n = k \). Let's assume this holds:\[ 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \cdots + k \cdot 2^k = 2[1 + (k-1)2^k] \]This assumption is not proven at this stage but serves as a pivotal step to move toward the next crucial step, the inductive step.

• By assuming for \( n = k \), we have a reference point for the subsequent proof step.
• This assumption is temporary and only serves to link our inductive process.

Inductive Step

The inductive step is where the magic happens. Here, you demonstrate that if the formula holds for \( n = k \), then automatically, it also holds for \( n = k + 1 \). You take advantage of your inductive hypothesis and extend it. Start from the formula's left-hand side for \( n = k + 1 \):\[ 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \cdots + k \cdot 2^k + (k+1) \cdot 2^{k+1} \]Incorporate the inductive hypothesis:\[ = 2[1 + (k-1)2^k] + (k+1) \cdot 2^{k+1} \]Step through the algebra to simplify:\[ = 2 + 2k \cdot 2^k + (k+1) \cdot 2^{k+1} \]Combine terms to show equality:\[ = 2 + (k+1) \cdot 2^{k+1} = 2[1 + k\cdot 2^{k+1}] \]Thus, you establish that if true for \( n = k \), then true for \( n = k + 1 \), sealing the argument.

• This step demonstrates the logical link from \( n = k \) to \( n = k+1 \).
• It confirms the formula's universal validity across all natural numbers.