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 Fundamentals
Prove \( 2+4+6+\cdots+2n = n(n+1) \) by mathematical induction.
Step 1: Show it is true for \( n=1 \).
LHS \( = 2 \times 1 = 2 \)
RHS \( = 1 \times ( 1+1) = 2 \)
LHS \( = \) RHS
It is true for \( n = 1 \).
Step 2: Assume that it is true for \( n=k \).
That is, \( 2+4+6+\cdots+2k = k(k+1) \).
Step 3: Show it is true for \( n = k+1 \).
That is, \( 2+4+6+\cdots+2k+2(k+1) = (k+1)(k+2) \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\text{LHS } &= 2+4+6+\cdots+2k+2(k+1) \\
&= k(k+1) + 2(k+1) &\color{green} \text{replaced by the assumption in Step 2}\\
&= (k+1)(k+2) &\color{green} \text{factorise by } (k+1) \\
&= \text{ RHS}
\end{aligned} \)
Therefore it is true for \( n=k+1 \).
Therefore the statement is true for \( n\ge 1 \).
Mathematical Induction with Indices
Prove \( 1 \times 2 + 2 \times 2^2 + 3 \times 2^3 + \cdots + n \times 2^n = (n-1) \times 2^{n+1} + 2 \) by mathematical induction.
Step 1: Show it is true for \( n=1 \).
LHS \( = 1 \times 2 = 2 \)
RHS \( = (1-1) \times 2^{1-1} + 2 = 2 \)
LHS \( = \) RHS
Therefore it is true for \( n=1 \).
Step 2: Assume that it is true for \( n=k \).
That is, \( 1 \times 2 + 2 \times 2^2 + 3 \times 2^3 + \cdots + k \times 2^k = (k-1) \times 2^{k+1} + 2 \).
Step 3: Show it is true for \( n=k+1 \).
That is \( 1 \times 2 + 2 \times 2^2 + 3 \times 2^3 + \cdots + k \times 2^k + (k+1) \times 2^{k+1} = k \times 2^{k+2} + 2 \)
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\text{LHS } &= 1 \times 2 + 2 \times 2^2 + 3 \times 2^3 + \cdots + k \times 2^k + (k+1) \times 2^{k+1} \\
&= (k-1) \times 2^{k+1} + 2 + (k+1) \times 2^{k+1} &\color{green} \text{replaced by the assumption in Step 2} \\
&= [(k-1)+(k+1)] \times 2^{k+1} + 2 &\color{green} \text{factorise by } 2^{k+1} \\
&= 2k \times 2^{k+1} + 2 \\
&= k \times 2^{k+2} + 2 \\
&= \text{RHS}
\end{aligned} \)
Therefore it is true for \( n=k+1 \).
Therefore the statement is true for \( n \ge 1 \).
Mathematical Induction with Factorials
Prove \( 2 \times 1! + 5 \times 2! + 10 \times 3! + \cdots + (n^2+1)n! = n(n+1)! \) by mathematical induction.
Note that \( \require{AMSsymbols} \require{color} \color{red} (n+1)n! = (n+1)! \)
Step 1: Show it is true for \( n=1 \).
LHS \( = (1^2+1) \times 1! = 2 \)
RHS \( = 1 \times (1+1)! = 2 \)
LHS \( = \) RHS
It is true for \( n = 1 \).
Step 2: Assume that it is true for \( n=k \).
That is, \( 2 \times 1! + 5 \times 2! + 10 \times 3! + \cdots + (k^2+1)k! = k(k+1)! \).
Step 3: Show it is true for \( n=k+1 \).
That is, \( 2 \times 1! + 5 \times 2! + 10 \times 3! + \cdots + (k^2+1)k!+ (k^2+2k+2)(k+1)! = (k+1)(k+2)! \)
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\text{LHS } &= 2 \times 1! + 5 \times 2! + 10 \times 3! + \cdots + (k^2+1)k!+ (k^2+2k+2)(k+1)! \\
&= k(k+1)!+ (k^2+2k+2)(k+1)! &\color{green} \text{replaced by the assumption in Step 2}\\
&= [k + (k^2+2k+2)](k+1)! &\color{green} \text{factorise by } (k+1)! \\
&= (k^2 + 3k + 2)(k+1)! \\
&= (k+1)(k+2)(k+1)! \\
&= (k+1)(k+2)! &\color{green} (k+2)(k+1)! = (k+2)! \\
&= \text{RHS}
\end{aligned} \)
Therefore it is true for \( n=k+1 \).
Therefore the statement is true for \( n \ge 1 \).
Mathematical Induction with Sigma \( \displaystyle \sum \) Notations
Prove \( \displaystyle \sum_{a=1}^{n} a^2 = \frac{1}{6}n(n+1)(2n+1) \) by mathematical induction.
Before proving the statement in sigma notation, it is highly recommended to expand to series format to avoid silly mistakes.
\( \displaystyle \sum_{a=1}^{n} a^2 = 1^2 + 2^2 + 3^2 + \cdots + n^2 \)
So the question becomes now as follows.
Prove \( \displaystyle 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{1}{6}n(n+1)(2n+1) \) by mathematical induction.
Step 1: Show it is true for \( n=1 \).
LHS \( = 1^2 = 1 \)
RHS \( = \frac{1}{6} \times 1 \times (1+1) \times (2+1) = 1 \)
LHS \( = \) RHS
It is true for \( n = 1 \).
Step 2: Assume that it is true for \( n=k \).
That is, \( \displaystyle 1^2 + 2^2 + 3^2 + \cdots + k^2 = \frac{1}{6}k(k+1)(2k+1) \).
Step 3: Show it is true for \( n=k+1 \).
That is, \( \displaystyle 1^2 + 2^2 + 3^2 + \cdots + k^2 + (k+1)^2 = \frac{1}{6}(k+1)(k+2)(2k+3) \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\text{LHS } &= 1^2 + 2^2 + 3^2 + \cdots + k^2 + (k+1)^2 \\
&= \frac{1}{6}k(k+1)(2k+1) + (k+1)^2 &\color{green} \text{replaced by the assumption in Step 2} \\
&= (k+1)\Big[\frac{1}{6}k(2k+1) + (k+1)\Big] &\color{green} \text{factorised by (k+1)} \\
&= (k+1) \times \frac{2k^2 + k + 6k + 6}{6} \\
&= \frac{1}{6}(k+1)(2k^2 + 7k + 6) \\
&= \frac{1}{6}(k+1)(k+2)(2k+3) \\
&= \text{RHS}
\end{aligned} \)
Therefore it is true for \( n=k+1 \).
Therefore the statement is true for \( n \ge 1 \).
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