Question

Prove by induction: (a) \(0 \cdot 2^{0}+1 \cdot 2^{1}+2 \cdot 2^{2}+3 \cdot 2^{3}+\cdots+n \cdot 2^{n}=(n-1) 2^{n+1}+2\) for \(n \geq 0\) (b) \(1^{2}+3^{2}+5^{2}+\cdots+(2 n+1)^{2}=(n+1)(2 n+1)(2 n+3) / 3\) for \(n \geq 0\) (c) \(1^{2}-2^{2}+3^{2}+\cdots+(-1)^{n-1} n^{2}=(-1)^{n-1} n(n+1) / 2\) for \(n \geq 0\) (d) \(1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n \cdot(n+1)=n(n+1)(n+2) / 3\) for \(n \geq 0\) (e) \(1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\cdots+n \cdot(n+1) \cdot(n+2)=n(n+1)(n+2)$$(n+3) / 4\) for \(n \geq 0\)

Short answer

Each statement can be proven true for all \(n\) using mathematical induction, validating at \(n=0\) and \(n=k+1\).

Step-by-step solution

Understand the base case (a)

We are tasked to prove that the formula is valid for \( n = 0 \). Substitute \( n = 0 \) in the formula to check if both sides are equal: \ \( 0 \cdot 2^0 = -1 \cdot 2^1 + 2 \) results in \( 0 = 0 \). The base case holds true.

Inductive step assumption (a)

Assume the formula \( 0 \cdot 2^{0}+1 \cdot 2^{1}+ ext{...}+k \cdot 2^{k}=(k-1) 2^{k+1}+2 \) is true for \( n = k \). Under this assumption, it follows: \ \( P(k) = (k-1)2^{k+1}+2 \).

Prove for n = k+1 (a)

Show that the formula is true for \( n = k+1 \) based on the assumption of \( P(k) \). \ Consider the left-hand side: \ \[ 0 \cdot 2^0 + 1 \cdot 2^1 + \cdots + k \cdot 2^k + (k+1) \cdot 2^{k+1} = (k-1) 2^{k+1} + 2 + (k+1) \cdot 2^{k+1} \]. \ Simplify the expression: \ \[= (k-1+1) 2^{k+2} + 2 = k \cdot 2^{k+2} + 2 \]. Thus, it matches the right side.

Verification complete for (a)

With the base case and inductive step verified, our proof for part (a) using mathematical induction is complete.

Base case for (b)

For part (b), substitute \( n = 0 \) in the given expression: \( (2 \times 0 + 1)^2 = 1 \). Verify this equates to the right side of the formula: \((0+1)(1)(3)/3 = 1 \). The base case holds.

Inductive hypothesis (b)

Assume \( 1^2 + 3^2 + \cdots + (2k+1)^2 = (k+1)(2k+1)(2k+3)/3 \) holds true for \( n = k \).

Prove for n = k+1 (b)

Prove it for \( n = k+1 \). Consider \( 1^2 + 3^2 + \cdots + (2k+1)^2 + (2(k+1)+1)^2 \).\ Now simplify: \( P(k) + (2k+3)^2 \). \ Insert our hypothesis expression and simplify to match the form of \( (k+2)(2k+3)(2k+5)/3 \).

Verification complete for (b)

Both base case and inductive step for part (b) are validated by matching expressions at \( n = k+1 \).

Base case for (c)

Check \( n = 0 \): \( \text{Left-hand side: } 0 \rightarrow \text{ Right-hand: } 0 \). The formula holds true.

Inductive hypothesis (c)

Assume true for \( n = k \): \\( 1^2 - 2^2 + 3^2 - \cdots + (-1)^{k-1} k^2 = (-1)^{k-1} \frac{k(k+1)}{2} \).

Prove for n = k+1 (c)

Check \( n = k+1 \): \Add term \( (-1)^k (k+1)^2 \): \Modify our hypothesis expression with this new term and simplify to match the form of \( \frac{-(k+1)(k+2)}{2} \).

Verification complete for (c)

Base case confirmed and inductive step reaffirmed for part (c), therefore, the induction proof is complete.

Base case for (d)

Verify for \( n = 0 \): \Left-hand side is 0, matching \( n(n+1)(n+2)/3 = 0 \).

Inductive hypothesis (d)

Suppose it holds for \( n = k \): \( 1 \cdot 2 + 2 \cdot 3 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3} \).

Prove for n = k+1 (d)

Show that adding \( (k+1)(k+2) \) gives \( P(k) + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3} \). Use algebra to align terms appropriately.

Verification complete for (d)

Both components of the induction proof are satisfied: base case and incremental step at \( n = k+1 \).

Base case for (e)

For \( n = 0 \): Left-hand side is 0, aligning with \( n(n+1)(n+2)(n+3)/4 = 0 \).

Inductive hypothesis (e)

Assume valid for \( n = k \): \ \( 1 \cdot 2 \cdot 3 + \cdots + k(k+1)(k+2) = \frac{k(k+1)(k+2)(k+3)}{4} \).

Prove for n = k+1 (e)

Analyze adding \( (k+1)(k+2)(k+3) \): \Combining it with hypothesis should simplify to match expression \( \frac{(k+1)(k+2)(k+3)(k+4)}{4} \).

Verification complete for (e)

Ensure congruity for both the base case and extension \( n=k+1 \); proof is then finalized and verified for this fragment.

Key concepts

Understanding the Base Case

Mathematical induction begins with verifying the base case. The base case serves as the foundation of the induction process. It involves proving that a statement holds true for the initial value, usually specified as \( n = 0 \) or \( n = 1 \), depending on the problem's domain.
For instance, in the problem of proving the formula \( 0 \cdot 2^0 + 1 \cdot 2^1 + 2 \cdot 2^2 + \cdots + n \cdot 2^n = (n-1) 2^{n+1} + 2 \) for \( n \geq 0 \), the base case checks this expression when \( n = 0 \).
Here, substituting \( n = 0 \) results in both sides equating as \( 0 = 0 \), hence verifying the base case.
Concisely, the base case ensures that the property holds for the starting point of the mathematical narrative. Without this, the induction process cannot proceed.

The Inductive Hypothesis

The inductive hypothesis is a pivotal step in the induction process. It involves assuming that the given statement or property holds for an arbitrary value \( n = k \).
This assumption is crucial because it forms the bridge between the base case and the inductive step.
For example, assume that the expression \( 0 \cdot 2^0 + 1 \cdot 2^1 + \cdots + k \cdot 2^k = (k-1) 2^{k+1} + 2 \) is true for \( n = k \).
This assumption is not about proving the exact value at this moment but rather setting up a logical framework that allows the next step to transform and prove consistency as \( n \) advances.

• Start with a solid assumption for a generic \( k \).
• Use simple algebraic manipulation to simplify expressions.

This step requires a firm trust that if the hypothesis holds, you can leverage it to validate the subsequent case in the inductive step.

Transitioning with the Inductive Step

The inductive step is where the magic of mathematical induction unfolds. It completes the chain by proving that if the statement holds for \( n = k \), it must also hold for \( n = k+1 \).
In this phase, you'll start with the equation assumed true in the inductive hypothesis.
Using algebraic transformations and logical reasoning, you will attempt to show that adding the next element transforms to fit the asserted formula at the next stage.
For example, you enhance the expression \( 0 \cdot 2^0 + 1 \cdot 2^1 + \cdots + k \cdot 2^k + (k+1) \cdot 2^{k+1} \) and demonstrate that the right side maintains its integrity, transforming to \( (k+1)2^{k+2} + 2 \), confirming the pattern continues.

• Carefully add increments to expressions.
• Simplify to match the pattern of the desired formula.

Essentially, the inductive step ensures the formula holds at every subsequent step, creating a domino effect from your base case up through any desired \( n \).