Use mathematical induction to prove that each statement is true for every positive integer n. 1/(1 · 2) + 1/(2 · 3) + 1/(3 · 4) + ... + 1/(n(n+1)) = n/(n + 1)

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Start by defining the statement to prove using mathematical induction. Let \( P(n) \) be the statement: \[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1} \].

Check the base case \( n=1 \). Substitute \( n=1 \) into the left-hand side and right-hand side of the equation to verify that both sides are equal.

Assume the statement \( P(k) \) is true for some positive integer \( k \), that is, assume \[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1} \]. This is the induction hypothesis.

Use the induction hypothesis to prove \( P(k+1) \) is true. Add the next term \( \frac{1}{(k+1)(k+2)} \) to both sides of the assumed equation and simplify the right-hand side to show it equals \( \frac{k+1}{k+2} \).

Conclude that since the base case is true and the induction step holds, by mathematical induction, the statement is true for every positive integer \( n \).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Mathematical Induction

Mathematical induction is a proof technique used to establish that a statement holds for all positive integers. It involves two steps: proving the base case (usually n=1) is true, and then proving that if the statement holds for an arbitrary integer k, it also holds for k+1. This creates a chain of truth for all n.

Telescoping Series

A telescoping series is a sum where many terms cancel out when expanded, simplifying the expression significantly. In this problem, the sum of fractions 1/(k(k+1)) can be decomposed into partial fractions, allowing terms to cancel and making it easier to find a closed form for the sum.

Partial Fraction Decomposition

Partial fraction decomposition breaks a complex rational expression into simpler fractions that are easier to sum or integrate. For example, 1/(k(k+1)) can be written as 1/k - 1/(k+1), which helps in simplifying the sum and proving the formula using induction.