Proving the Formula 1x3 2x4 3x5 … nn^2 n(n 1)(2n 7)/6 Using Mathematical Induction

Mathematical induction is a powerful technique for proving formulas in discrete mathematics. In this article, we will use mathematical induction to prove the formula for the sum of the series 1x3 2x4 3x5 … nn^2. The formula is:

Series and Formula

Let's define the series we want to prove the formula for:

1times;3 2times;4 3times;5 u2026 ntimes;(n 2)

We will show that the sum of this series is given by the formula:

n(n 1)(2n 7)/6

Step-by-Step Proof

Step 1: Base Case (n 1)

First, we check the base case where n 1.

The left-hand side (LHS) for n 1 is:

1times;3 3

The right-hand side (RHS) for n 1 is:

1(1 1)(2(1) 7)/6 1times;2times;9/6 18/6 3

Both sides are equal, so the base case holds.

Step 2: Inductive Step

Assume the statement is true for some k.

1times;3 2times;4 3times;5 u2026 ktimes;(k 2) k(k 1)(2k 7)/6

We need to prove it for k 1. The sum for n k 1 is:

1times;3 2times;4 3times;5 u2026 ktimes;(k 2) (k 1times;(k 3))

According to the inductive hypothesis:

k(k 1)(2k 7)/6 (k 1)(k 3)

We will simplify the right-hand side:

Step 3: Simplifying the Right-Hand Side

We factor out (k 1):

(k 1)[(k)(2k 7)/6 (k 3)]

Now, we combine the terms inside the brackets:

(k 1)[(2k^2 7k)/6 (k 3)] (k 1)[(2k^2 13k 18)/6]

Step 4: Factor the Quadratic

The quadratic expression 2k^2 13k 18 can be factored as:

2k^2 13k 18 (2k 9)(k 2)

Substitute this back into the expression:

(k 1) [(2k 9)(k 2)/6] (k 1) [2k^2 13k 18]/6

Now we check if the resulting expression matches the formula:

(k 1) [2k^2 13k 18]/6 (k 1) [2k^2 2k k 18]/6 (k 1) [2k(k 1) (k 18)]/6

Which simplifies to:

(k 1) [2k(k 1) 18 k]/6 (k 1) [2k(k 1) (k 18)]/6 (k 1) [2k(k 1) (k 18)]/6 (k 1) [2k(k 1) (k 18)]/6 (k 1) [2k(k 1) (k 18)]/6

This can be further simplified to:

(k 1) [2k^2 7k 18 k]/6 (k 1) [2k^2 8k 18]/6 (k 1) [2k(k 1) 18]/6 (k 1) [2k(k 1) (k 18)]/6

Thus, we have:

(k 1) [2k(k 1) (k 18)]/6 [2(k 1)((k 1) 1) (2(k 1) 7)]/6

Step 5: Conclusion

By the principle of mathematical induction, the original statement is true for all n ≥ 1.

Simplified Formula Derivation

It's also useful to derive the sum of the series sum;_1^n k(k 2) using a different approach.

The series can be expressed as:

sum;_1^n k(k 2) sum;_1^n k^2 2sum;_1^n k

This can be broken down using known summation formulas:

sum;_1^n k^2 n(n 1)(2n 1)/6

sum;_1^n k n(n 1)/2

Therefore:

sum;_1^n k(k 2) n(n 1)(2n 1)/6 2n(n 1)/2 n(n 1)(2n 7)/6

Which matches our original formula.