Mathematical Induction: Proving Summation Formulas
Initial Problem: Volume and Surface Area Equation
Given Information:
The link (length) of an object is twice its width (L = 2W).
The total surface area of the exterior walls is 5000 ext{ ft}^2 for a rectangular shape.
Formulating Surface Area Equation:
The four walls consist of two front/back walls and two side walls.
Front/back walls: Each has area L imes H. Since there are two, this is 2LH.
Side walls: Each has area W imes H. Since there are two, this is 2WH.
Total surface area: 2WH + 2LH = 5000.
Substitution and Simplification:
Substitute L = 2W into the surface area equation: 2WH + 2(2W)H = 5000.
This simplifies to 2WH + 4WH = 5000.
Combined: 6WH = 5000.
Solving for Height (H):
Divide by 6W: H = rac{5000}{6W}.
Volume Equation in terms of Width (W):
The volume of a rectangular prism is V = LWH.
Substitute L = 2W: V(W) = (2W)WH = 2W^2H.
Substitute the expression for H: V(W) = 2W^2 imes rac{5000}{6W}.
Simplify the volume equation: V(W) = rac{10000W^2}{6W} = rac{5000W}{3}.
Review of Summation Formulas
Sum of first n integers: \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
Sum of first n squares: \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}
Sum of first n cubes: \sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2
Mathematical Induction: Core Analogy and Process
Analogy: Knocking down a line of dominos.
Base Case: The first domino (n=1) falls.
Induction Hypothesis/Step: If any domino k falls, then the next domino (k+1) will also fall.
Purpose: Proving these summation formulas are true for all natural numbers n.
Proof by Induction: Sum of First n Integers
Statement to Prove: \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
1. Base Case (Show true for n=1):
Left Hand Side (LHS): \sum_{i=1}^{1} i = 1
Right Hand Side (RHS): \frac{1(1+1)}{2} = \frac{1 \times 2}{2} = 1
Conclusion: Since LHS = RHS, the statement is true for n=1. (The first domino falls.)
2. Induction Hypothesis (Assume true for some k < n):
Assume the statement is true for some integer k. Specifically, assume: \sum_{i=1}^{k} i = \frac{k(k+1)}{2}.
3. Induction Step (Show true for k+1):
Goal: We want to show that the statement is true for k+1, meaning: \sum_{i=1}^{k+1} i = \frac{(k+1)((k+1)+1)}{2} = \frac{(k+1)(k+2)}{2}.
Start with LHS: \sum_{i=1}^{k+1} i
Expand the sum: 1 + 2 + 3 + \dots + k + (k+1).
Rewrite using summation notation: \left(\sum_{i=1}^{k} i\right) + (k+1).
Apply Induction Hypothesis: Substitute the assumed formula for the sum up to k:
\frac{k(k+1)}{2} + (k+1).Find a common denominator:
\frac{k(k+1)}{2} + \frac{2(k+1)}{2} = \frac{k(k+1) + 2(k+1)}{2}.Factor out the common term (k+1):
\frac{(k+1)(k+2)}{2}.Result: This matches the RHS that we aimed for, so the statement holds for k+1.
4. Conclusion (By mathematical induction):
By mathematical induction, the sum of i from 1 to n is equal to \frac{n(n+1)}{2} for all natural numbers n.
Proof by Induction: Sum of First n Squares
Statement to Prove: \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}
1. Base Case (Show true for n=1):
LHS: \sum_{i=1}^{1} i^2 = 1^2 = 1
RHS: \frac{1(1+1)(2(1)+1)}{6} = \frac{1 \times 2 \times 3}{6} = 1
Conclusion: Since LHS = RHS, the statement is true for n=1. The base case holds.
2. Induction Hypothesis (Assume true for some k < n):
Assume the statement is true for some integer k. Specifically, assume: \sum_{i=1}^{k} i^2 = \frac{k(k+1)(2k+1)}{6}.
3. Induction Step (Show true for k+1):
Goal: We want to show that the statement is true for k+1, meaning: \sum_{i=1}^{k+1} i^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}.
Start with LHS: \sum_{i=1}^{k+1} i^2
Rewrite using summation notation: \left(\sum_{i=1}^{k} i^2\right) + (k+1)^2.
Apply Induction Hypothesis: Substitute the assumed formula for the sum up to k:
\frac{k(k+1)(2k+1)}{6} + (k+1)^2.Find a common denominator: Force a common denominator of 6:
\frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6} = \frac{k(k+1)(2k+1) + 6(k+1)^2}{6}.Factor out the common term (k+1):
\frac{(k+1)[k(2k+1) + 6(k+1)]}{6}.Simplify the expression inside the bracket: Distribute and combine like terms:
\frac{(k+1)[2k^2 + k + 6k + 6]}{6} = \frac{(k+1)(2k^2 + 7k + 6)}{6}.Factor the quadratic term (2k^2 + 7k + 6): This factors into (2k+3)(k+2).
(Self-check: (2k+3)(k+2) = 2k^2 + 4k + 3k + 6 = 2k^2 + 7k + 6)Substitute the factored quadratic back:
\frac{(k+1)(k+2)(2k+3)}{6}.Result: This matches the RHS that we aimed for, so the statement holds for k+1.
4. Conclusion (By mathematical induction):
By mathematical induction, the sum of i^2 from 1 to n is equal to \frac{n(n+1)(2n+1)}{6} for all natural numbers n.
Grading an Induction Proof
An induction proof is considered a "mathematical essay" requiring precise language and organized steps.
Typical Point Breakdown (e.g., for a 15-point problem):
Base Case: 3 points (Statement, calculation of LHS/RHS, conclusion).
Induction Hypothesis: 2 points (Clearly stating the assumption for k).
Induction Step Statement: 1 point (Clearly stating the goal to prove for k+1).
Algebraic Work in Induction Step: 7 points (Demonstrating the transformation from the k-sum to the k+1 sum).
Conclusion: 2 points (Formal statement by mathematical induction).
Key Emphasis: The language in each section (e.g., "Since LHS = RHS, the statement is true for n=1") is crucial for a complete and correct proof.
Algebraic steps must be shown, and simplification shortcuts are accepted if algebraically correct.
Connection to Previous Lectures
These summation formulas are directly applicable to problems encountered earlier, particularly when calculating the area under a curve using Riemann sums with n rectangles. Previously, we derived expressions like the sum of i^2 from 1 to n when setting up these problems.
Now, instead of leaving the sum in that form, we can substitute the derived closed-form formulas (e.g., \frac{n(n+1)(2n+1)}{6} for \sum i^2) to simplify calculations and find a direct formula in terms of n.