Tag Archives: Factorials

Mathematical Induction Regarding Factorials

Mathematical Induction Regarding Factorials

Prove by mathematical induction that for all integers \( n \ge 1 \) ,$$ \dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \cdots + \dfrac{n}{(n+1)!} = 1-\dfrac{1}{(n+1)!}$$ Step 1 Show it is true for \( n=1 \).\( \begin{align} \displaystyle\text{LHS } &= \dfrac{1}{2!} = \dfrac{1}{2} \\\text{RHS } &= 1-\dfrac{1}{2!} \\&= 1-\dfrac{1}{2} \\&= \dfrac{1}{2} \end{align} \)Thus, the statement is […]

Binomial Coefficient

Binomial Coefficient

$$\binom{n}{k}=\dfrac{n!}{k!(n-k)!}$$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 binomial coefficients […]