How to Prove by Induction That 1/2 × 1/6 × ... × 1/{n(n-1)} n/(n-1)

Mathematical induction is a powerful tool in mathematics used to prove statements for all natural numbers. In this article, we will show how to use mathematical induction to prove that the product of the series 1/2 × 1/6 × ... × 1/{n(n-1)} n/(n-1).

Introduction to Mathematical Induction

Mathematical induction is a method of mathematical proof typically used to establish that a proposition is true for all natural numbers. The process involves two steps:

Base Case: Prove that the statement is true for the smallest value (usually n 1). Inductive Step: Assume that the statement is true for n k, and then prove that it is also true for n k 1.

This ensures that the statement is true for all natural numbers.

Proving the Given Statement

Consider the statement: fn 1/2 × 1/6 × ... × 1/{n(n-1)} n/(n-1).

Step 1: Base Case

For n 1:

Calculate 1/2: Substitute 1/2 for f1, which is equal to 1/(1-1 1) 1/1 1.

Thus, the statement is true for the base case n 1.

Step 2: Inductive Step

Assume the statement is true for n k and prove it for n k 1.

Start with the inductive hypothesis: fk k/(k-1). Consider the next term in the product, which is 1/{(k 1)k}. Multiply both sides of the inductive hypothesis by 1/{(k 1)k}: fk × 1/{(k 1)k} k/(k-1) × 1/{(k 1)k} Simplify the right-hand side: fk 1 [k/(k-1)] × 1/{(k 1)k} {k/(k-1)} × 1/{(k 1)k} Further simplify: fk 1 k/(k^2-k) × 1/{(k 1)k} (k/(k^2-k)) × (1/{(k 1)k}) Simplify the expression: fk 1 (k/(k^2-k)) × (1/{(k 1)k}) {k/(k^2-k)} × (1/{(k 1)k}) Combine terms in the numerator and the denominator: fk 1 (k/(k^2-k)) × (1/{(k 1)k}) (k/(k-1)) × (1/{(k 1)k}) Simplify further: fk 1 (k/(k-1)) × (1/(k 1)k) (k/(k-1)) × (1/(k(k 1))) Combine terms in the numerator and denominator: fk 1 (k/(kk-1)) × (1/(k 1)) (k/(kk-1)) × (1/(k 1)) (k/(k^2-k)) × (1/(k 1)) Finally, simplify: fk 1 (k/(k^2-k)) × (1/(k 1)) (k/(k^2-k)) × (1/(k 1)) (k/(k^2-k)) × (1/(k 1)) (k/(k-1)) × (1/(k 1)) (k 1)/(k 1-1) (k 1)/k

Hence, the statement is true for n k 1, and by the principle of mathematical induction, it is true for all n ∈ ?.

Conclusion

By successfully demonstrating the base case and the inductive step, we have proven that the product of the series 1/2 × 1/6 × ... × 1/{n(n-1)} n/(n-1) for all natural numbers n.

Additional Resources

For further reading and practice, explore comprehensive resources on mathematical induction and series proof techniques. These will help you develop a deeper understanding and enhance your skills in proving statements using induction.

Frequently Asked Questions

What is the difference between base case and inductive step in mathematical induction?

The base case establishes the truth of the statement for the smallest value of n, while the inductive step shows that if the statement is true for n k, it is also true for n k 1. Together, they ensure the statement is true for all natural numbers.

How can I apply induction to more complex sequences?

The process remains the same, but the expressions might be more complex. The key is to break down the problem into simpler parts and use the inductive hypothesis to prove the next step.