# Tag Archives: Factorials 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!} \\ […] # Binomial Coefficient \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 […] # Mathematical Induction Fundamentals

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 […] # Mathematical Induction Inequality Proof with Factorials

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 […] # Mathematical Induction Inequality Proof with Two Initials

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 […]