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