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Chapter 5: Problem 14

Use mathematical induction in Exercises \(3-17\) to prove summation formulae. Be sure to identify where you use the inductive hypothesis. Prove that for every positive integer \(n, \sum_{k=1}^{n} k 2^{k}=\) \((n-1) 2^{n+1}+2\).

Short Answer

Expert verified
\(\sum_{k=1}^{n} k 2^{k} = (n-1) 2^{n+1} + 2\). It is proven by induction.

Step by step solution

01

Understand the Problem

The goal is to prove that for every positive integer \(n\), the formula \(\sum_{k=1}^{n} k 2^{k} = (n-1) 2^{n+1} + 2\) holds true using mathematical induction.
02

Base Case

Verify the formula for the smallest positive integer, \(n = 1\). Compute both sides of the equation separately: Left-hand side: \(\sum_{k=1}^{1} k 2^{k} = 1 \cdot 2^{1} = 2\) Right-hand side: \((1-1)2^{1+1} + 2 = 0 \cdot 4 + 2 = 2\) Since both sides are equal, the base case is verified.
03

Inductive Hypothesis

Assume that the formula holds for some positive integer \(n = k\). That is, assume \(\sum_{k=1}^{k} k 2^{k} = (k-1) 2^{k+1} + 2\).
04

Inductive Step

Show that if the formula holds for \(n=k\), it must also hold for \(n=k+1\). Consider: \(\sum_{k=1}^{k+1} k 2^{k} = \left(\sum_{k=1}^{k} k 2^{k}\right) + (k+1)2^{k+1}\) Using the inductive hypothesis: \((k-1) 2^{k+1} + 2 + (k+1)2^{k+1}\) Combine like terms: \((k-1)2^{k+1} + (k+1)2^{k+1} + 2\) \((k-1 + k+1) 2^{k+1} + 2\) \(2k 2^{k+1} + 2\) \((k+1-1)2^{(k+1)+1} + 2\)
05

Conclusion

Since it has been shown that if the formula holds for \(n=k\), it must also hold for \(n=k+1\), by mathematical induction, the formula \(\sum_{k=1}^{n} k 2^{k} = (n-1) 2^{n+1} + 2\) is true for all positive integers \(n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

summation formula
In mathematics, a summation formula provides a concise way to represent the sum of a sequence of numbers. For example, the formula we aim to prove here is \(\sum_{k=1}^{n} k 2^{k} = (n-1) 2^{n+1} + 2\).
Summation formulas are pivotal in various fields like physics, computer science, and probability. They help us quickly and accurately calculate results of adding many numbers without carrying out each operation manually.
Understanding these formulas lets us tackle more complex problems efficiently.
base case
The base case is the first step in mathematical induction, where we verify that the given formula works for the initial value of the variable. For instance, here we start with \(n = 1\).
The base case confirmation ensures our formula holds true at the beginning of the sequence.
For \(n = 1\), both sides of the equation \(\sum_{k=1}^{1} k 2^{k} = 2\) match, establishing a foundation for our inductive proof.
If the base case is not true, our formula does not hold for any greater values of \(n\).
inductive hypothesis
The inductive hypothesis is where we assume the formula is true for some positive integer \(k\). This assumption helps us to build a bridge from known to unknown values.
Here, we assume that \(\sum_{k=1}^{k} k 2^{k} = (k-1) 2^{k+1} + 2\) holds true for \(n = k\). This assumption is crucial as it sets up the framework for the next step.
By using this hypothesis, we can verify the formula for the next integer, moving step by step towards proving it for all positive integers.
inductive step
The inductive step is the final piece of the induction process. Here, we need to prove that if the formula holds for \(n = k\), it must also hold for the next integer \(n = k + 1\).
Consider: \(\sum_{k=1}^{k+1} k 2^{k} = (\sum_{k=1}^{k} k 2^{k}) + (k+1)2^{k+1}\).
Using our inductive hypothesis: \( (k-1) 2^{k+1} + 2 + (k+1)2^{k+1} \), we combine like terms to get: \( (k-1)2^{k+1} + (k+1)2^{k+1} + 2 \), which simplifies to \(2k 2^{k+1} + 2 \), confirming our formula for \(n = k + 1\).
This process proves our formula holds for all positive integers, completing the proof by induction.

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Most popular questions from this chapter

Exercises 22 and 23 present examples that show inductive loading can be used to prove results in computational geometry. Let \(P(n)\) be the statement that when non intersecting diagonals are drawn inside a convex polygon with \(n\) sides, at least two vertices of the polygon are not endpoints of any of these diagonals. a) Show that when we attempt to prove \(P(n)\) for all integers \(n\) with \(n \geq 3\) using strong induction, the in- ductive step does not go through. b) Show that we can prove that \(P(n)\) is true for all inte- gers \(n\) with \(n \geq 3\) by proving by strong induction the stronger assertion \(Q(n),\) for \(n \geq 4,\) where \(Q(n)\) states that whenever nonintersecting diagonals are drawn inside a convex polygon with \(n\) sides, at least two nonadjacent vertices are not endpoints of any of these diagonals.

What is wrong with this "proof" by strong induction? "Theorem" For every nonnegative integer \(n, 5 n=0\) . Basis Step: \(5 \cdot 0=0\) Inductive Step: Suppose that \(5 j=0\) for all nonnegative integers \(j\) with \(0 \leq j \leq k .\) Write \(k+1=i+j\) , where \(i\) and \(j\) are natural numbers less than \(k+1 .\) By the inductive hypothesis, \(5(k+1)=5(i+j)=5 i+5 j=0+0=0\)

When does a string belong to the set \(A\) of bit strings defined recursively by $$ \begin{array}{l}{\lambda \in A} \\ {0 x 1 \in A \text { if } x \in A}\end{array} $$ where \(\lambda\) is the empty string?

Let \(b\) be a fixed integer and \(j\) a fixed positive integer. Show that if \(P(b), P(b+1), \ldots, P(b+j)\) are true and \([P(b) \wedge P(b+1) \wedge \cdots \wedge P(k)] \rightarrow P(k+1)\) is true for every integer \(k \geq b+j,\) then \(P(n)\) is true for all integers \(n\) with \(n \geq b\) .

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