Worked Example
Prove that \( (2n)! > 2^n (n!)^2 \) using mathematical induction for \(n \ge 2 \).
Step 1: Show it is true for \( n =2 \).
\( \begin{aligned} \require{AMSsymbols} \require{color}
\text{LHS } &= (2 \times 2)! = 16 \\
\text{RHS } &= 2^2 \times (2!) = 8 \\
\text{LHS } &> { RHS} \\
\end{aligned} \)
\( \therefore \text{It is true for } n =2 \)
Step 2: Assume it is true for \(n =k\), that is \( (2k)! > 2^k (k!)^2. \)
Step 3: Show it is true for \(n =k+1\), that is \( (2k+2)! > 2^{k+1} \big[(k+1)!\big]^2. \)
\( \begin{aligned} \require{color}
\text{LHS } &= (2k+2)! \\
&= (2k+2)(2k+1)(2k)! \\
&= 2(k+1)(2k+1)(2k)! \\
&> 2(k+1)(2k+1)2^k (k!)^2 &\color{red} \text{by assumption} \\
&> (k+1)(2k+1)2^{k+1} (k!)^2 \\
&> (k+1)(k+1)2^{k+1} (k!)^2 &\color{red} 2k+1>k+1 \\
&= 2^{k+1} \big[(k+1)k!\big]^2 \\
&= 2^{k+1} \big[(k+1)!\big]^2 \\
&= \text{RHS}
\end{aligned} \)
\( \therefore (2k+2)! > 2^{k+1} \big[(k+1)!\big]^2 \text{ for } \ge 2.\)
Algebra Algebraic Fractions Arc Binomial Expansion Capacity Common Difference Common Ratio Differentiation Double-Angle Formula Equation Exponent Exponential Function Factorials Factorise Functions Geometric Sequence Geometric Series Index Laws Inequality Integration Kinematics Length Conversion Logarithm Logarithmic Functions Mass Conversion Mathematical Induction Measurement Perfect Square Perimeter Prime Factorisation Probability Product Rule Proof Quadratic Quadratic Factorise Ratio Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume
