Mathematical induction ：練習問題

When proving the statement \(P(n): \sum_{r=1}^n \frac{1}{(2r-1)(2r+1)} = \frac{n}{2n+1}\) for all positive integers \(n\) by mathematical induction, what is the value of the left-hand side of the base case \(P(1)\)?

Let \(P(n)\) be the statement \(\sum_{r=1}^n \frac{1}{(3r-2)(3r+1)} = \frac{n}{3n+1}\). In the inductive step of a proof by mathematical induction, we assume \(P(k)\) is true for some positive integer \(k\) and consider the sum for \(n = k+1\). Which of the following is the correct expression for the \((k+1)\)-th term that must be added to the sum of the first \(k\) terms?

Let \( P(n) \) be the statement \( \sum_{r=1}^n \frac{1}{(2r-1)(2r+1)(2r+3)} = \frac{n(n+2)}{3(2n+1)(2n+3)} \). During the proof of \( P(n) \) by mathematical induction, assuming \( P(k) \) is true, the Left-Hand Side (LHS) of \( P(k+1) \) becomes \( \frac{k(k+2)}{3(2k+1)(2k+3)} + T_{k+1} \). What is the expression for \( T_{k+1} \) and the simplified RHS of \( P(k+1) \)?

Let \(P(n)\) be the statement \(\sum_{r=1}^n r(r+3) = \frac{n(n+1)(n+5)}{3}\) for all positive integers \(n\). To prove this by mathematical induction, what are the values of the Left-hand Side (LHS) and the Right-hand Side (RHS) for the base case \(P(1)\)?

Consider the proposition \( P(n): \sum_{r=1}^n r(r+1)(r+2) = \frac{n(n+1)(n+2)(n+3)}{4} \). In the inductive step, if we assume \( P(k) \) is true for some positive integer \( k \), which of the following is the correct expression for the sum \( \sum_{r=1}^{k+1} r(r+1)(r+2) \) using the inductive hypothesis?

When proving by mathematical induction that $$P(n): 4^n + 5$$ is divisible by $$3$$ for all positive integers $$n$$, what is the value of $$P(1)$$?

For the proposition $$P(n): \sum_{i=1}^{n} i(i+1) = \frac{n(n+1)(n+2)}{3}$$. Assuming $$P(k)$$ is true, i.e., $$\sum_{i=1}^{k} i(i+1) = \frac{k(k+1)(k+2)}{3}$$. Simplify the expression for $$\sum_{i=1}^{k+1} i(i+1)$$ to show it matches the formula for $$n=k+1$$.

Prove by mathematical induction that for all positive integers \(n\),

$$\sum_{r=1}^{n} \frac{2r+1}{r^2(r+1)^2} = 1 - \frac{1}{(n+1)^2}$$

Prove by mathematical induction that for all positive integers $$n$$, the sum of the series $$1+5+9+\dots+(3n-1)$$ is given by $$\sum_{r=1}^{n} (3r-1) = \frac{n(3n+1)}{2}$$.

Consider the sequence \( a_n \) defined by \( a_n = \frac{1}{n(n+2)} \) for all positive integers \( n \).

(a) Find constants \( A \) and \( B \) such that \( a_n = \frac{A}{n} + \frac{B}{n+2} \).

(b) Let \( S_n = \sum_{k=1}^n a_k \). Prove by mathematical induction that \( S_n = \frac{n(3n+5)}{4(n+1)(n+2)} \) for all positive integers \( n \).