# Tag Archives: Factorials Prove $2 \times 1! + 5 \times 2! + 10 \times 3! + \cdots + (n^2+1)n! = n(n+1)!$. Step 1 Show it is true for $n=1$. \begin{align} &\text{LHS} = (1^2+1) \times 1! = 2 \\ &\text{RHS} = 1 \times (1+1)! = 2 \\ &\text{LHS} = \text{RHS} \\ &\text{Therefore it is […] # 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

$$\binom{n}{k}=\dfrac{n!}{k!(n-k)!}$$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 […] # Mathematical Induction Fundamentals

The Mathematical Induction Fundamentals are defined for applying 3 steps: 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 […] # 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{AMSsymbols} \require{color}\text{LHS } &= (2 \times 2)! = 16 \\\text{RHS } &= 2^2 \times (2!) = 8 \\\text{LHS } &> { RHS} \\\end{aligned}\( \therefore […] # Mathematical Induction Inequality Proof with Two Initials

Usually, mathematical induction inequality proof requires one initial value, but in some cases, two initials are required, such as in the Fibonacci sequence. In this case, it is necessary 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 third […]