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Prove by mathematical induction that for al lintegers \( n \ge 1 \) , $$ \dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \cdots + \dfrac{n}{(n+1)!} = 1 – \dfrac{1}{(n+1)!}$$ Step 1: Show it is true for \( n=1 \). \( \begin{align} \displaystyle \text{LHS } &= \dfrac{1}{2!} = \dfrac{1}{2} \\ \text{RHS } &= 1 – \dfrac{1}{2!} \\ […]
$$\binom{n}{k}=\dfrac{n!}{k!(n-k)!}$$Note that the binomial coefficient is sometimes written $^nC_k$ or $C^n_k$, depending on authors or geographical regions. \( \begin{aligned}\binom{n}{k} &= \dfrac{n!}{k!(n-k)!} \cdots (1) \\\binom{n}{n-k} &= \dfrac{n!}{(n-k)!(n-(n-k))!} = \dfrac{n!}{(n-k)!k!} \cdots (2) \\\therefore \binom{n}{k} &= \binom{n}{n-k} \text{by } (1) \text{ and } (2) \\\end{aligned} \) This means;\( \begin{aligned}\binom{10}{2} &= \binom{10}{8} \\\binom{100}{1} &= \binom{100}{99} \\\end{aligned} \) The following […]
$n!$ is the product of the first $n$ positive integers for $n\ge 1$.\( n!=n(n-1)(n-2)(n-3) \cdots \times 3 \times 2 \times 1 \)$n!$ is read “$n$ factorial”. For example, the product $5 \times 4 \times 3 \times 2 \times 1$ can be written as $5!$.Notice that $5 \times 4 \times 3$ can be written using factorial […]
The Mathematical Induction Fundamentals are defined for applying 3 steps, such as step 1 for showing its initial ignite, step 2 for making an assumption, and step 3 for showing it is true based on the assumption. Make sure the Mathematical Induction Fundamentals should be used only when the question asks to use it. Basic […]
Worked Example Prove that \( (2n)! > 2^n (n!)^2 \) using mathematical induction for \(n \ge 2 \). Step 1: Show it is true for \( n =2 \).\( \begin{aligned} \require{color}\text{LHS } &= (2 \times 2)! = 16 \\\text{RHS } &= 2^2 \times (2!) = 8 \\\text{LHS } &> { RHS} \\\end{aligned} \)\( \therefore \text{It […]
Usually, mathematical induction inequality proof requires one initial value, but in some cases, two initials are to be required, such as the Fibonacci sequence. In this case, it is required to show two initials are working as the first step of the mathematical induction inequality proof, and two assumptions are to be placed for the […]