FMAT pure1 series and induction

Question 1

What is the sum of the first n natural numbers?

Answer 1

The sum is given by:
Σr (from r = 1 to n) = n(n + 1) / 2

Question 2

How is the sum of the first n natural numbers derived using triangle numbers?

Answer 2

The nth triangle number T_n forms a triangular array of dots.
Two such triangles form a rectangle with rows n and columns n + 1.
T_n = n(n + 1) / 2

Question 3

What is the sum of the squares of the first n natural numbers?

Answer 3

The sum is given by:
Σr² (from r = 1 to n) = n(n + 1)(2n + 1) / 6

Question 4

Example: Find Σr² (from r = 1 to 50).

Answer 4

Σr² = (50 * 51 * 101) / 6 = 42925

Question 5

What is the sum of the cubes of the first n natural numbers?

Answer 5

The sum is given by:
Σr³ (from r = 1 to n) = [n(n + 1) / 2]²

Question 6

Example: Find Σr³ (from r = 1 to 10).

Answer 6

Σr³ = [(10 * 11) / 2]² = 55² = 3025

Question 7

What is the method of differences?

Answer 7

This method expresses a series as differences of terms, canceling intermediate terms to simplify the sum.

Question 8

Example: Simplify Σ (r(r + 1)) for r = 1 to n.

Answer 8

1. Expand r(r + 1) = r² + r.
2. Use standard results for Σr² and Σr:
   Σr(r + 1) = Σr² + Σr = [n(n + 1)(2n + 1) / 6] + [n(n + 1) / 2].

Question 9

How are partial fractions used to sum series?

Answer 9

Decompose a rational expression into partial fractions, then use the method of differences to simplify.

Question 10

Example: Sum Σ (2 / [(r + 1)(r + 2)]) for r = 1 to n.

Answer 10

1. Decompose: 2 / [(r + 1)(r + 2)] = 2 / (r + 1) - 2 / (r + 2).
2. Apply method of differences:
   Σ = 1 - 2/(n + 2).

Question 11

What is a convergent series?

Answer 11

A series is convergent if its sum approaches a finite limit as the number of terms approaches infinity.

Question 12

Example: Find the sum to infinity of Σ (2 / [(r + 1)(r + 2)]).

Answer 12

Using the sum for n terms: 1 - 2/(n + 2).
As n → ∞, 2/(n + 2) → 0.
Sum to infinity = 1

Question 13

How do you ensure correct cancellation in the method of differences?

Answer 13

Carefully identify and retain any remaining terms after cancellation, especially at the beginning and end of the series.

Question 14

What is proof by induction?

Answer 14

A mathematical proof method used to show that a statement is true for all positive integers.

Question 15

What are the three essential steps in proof by induction?

Answer 15

1. Prove the statement is true for a starting value (e.g., n = 1).
2. Assume it is true for n = k.
3. Prove it is true for n = k + 1.

Question 16

Example: Prove Σr (from r = 1 to n) = n(n + 1) / 2 using induction.

Answer 16

1. Base case (n = 1): LHS = 1; RHS = 1(1 + 1)/2 = 1. True.
2. Assume true for n = k: Σr = k(k + 1)/2.
3. For n = k + 1: Σr = [k(k + 1)/2] + (k + 1).
   Simplify to (k + 1)(k + 2)/2. True by induction.

Question 17

What is a counterexample in proofs?

Answer 17

A single example showing that a statement is false, disproving it.

Question 18

How is proof by induction used for divisibility?

Answer 18

1. Prove the base case satisfies divisibility.
2. Assume true for n = k.
3. Show that if true for n = k, it is true for n = k + 1.

Question 19

Example: Prove 2^(n+1) - 1 is divisible by 3 for n ≥ 1.

Answer 19

1. Base case (n = 1): 2^(1+1) - 1 = 3. Divisible by 3.
2. Assume true for n = k: 2^(k+1) - 1 = 3m.
3. For n = k + 1: 2^(k+2) - 1 = 2 * 2^(k+1) - 1.
   = 2(3m + 1) - 1 = 3(2m) + 1. True.

Question 20

How is proof by induction applied to matrices?

Answer 20

1. Prove the result holds for n = 1.
2. Assume it holds for n = k.
3. Show it holds for n = k + 1 by matrix multiplication.

Question 21

Example: Prove A^n = [[1, 2n], [0, 2^n]] using induction.

Answer 21

1. Base case (n = 1): A = [[1, 2], [0, 2]]. True.
2. Assume true for n = k: A^k = [[1, 2k], [0, 2^k]].
3. For n = k + 1: A^(k+1) = A^k * A.
   Verify: [[1, 2k + 2], [0, 2^(k+1)]] matches. True.

Question 22

How is proof by induction used for sequences?

Answer 22

1. Prove the sequence formula is true for the first term.
2. Assume it is true for the k-th term.
3. Prove it is true for the (k + 1)-th term using the sequence definition.

Question 23

Example: Prove u_n = 3(4^(n-1)) - 1 satisfies the recurrence u_(n+1) = 4u_n + 3.

Answer 23

1. Base case (n = 1): u_1 = 3(4^0) - 1 = 2. Matches.
2. Assume true for n = k: u_k = 3(4^(k-1)) - 1.
3. For n = k + 1: u_(k+1) = 4[3(4^(k-1)) - 1] + 3.
   Simplifies to 3(4^k) - 1. True.

Question 24

What are common errors in proof by induction?

Answer 24

1. Skipping the base case.
2. Incorrectly assuming n = k implies n = k + 1 without proof.
3. Failing to clearly state the inductive step conclusion.

Question 25

Why does proof by induction work?

Answer 25

It establishes a domino effect: proving the base case and the inductive step ensures the result holds for all positive integers.