IIT-JEE 1987Question 1
Prove by mathematical induction that $\dfrac{(2n)!}{2^{2n}(n!)^{2}}\le\dfrac{1}{(3n+1)^{1/2}}$ for all positive integers $n$.
Mathematical Reasoning
Answerproved
Solution
Let $u_n=\frac{(2n)!}{4^n(n!)^2}$. $u_{n+1}=u_n\frac{2n+1}{2n+2}$. If $u_n^2\le\frac1{3n+1}$ then $u_{n+1}^2\le\frac{(2n+1)^2}{(2n+2)^2(3n+1)}\le\frac1{3n+4}$ because $(2n+1)^2(3n+4)\le(2n+2)^2(3n+1)$ (i.e. $12n^3+28n^2+19n+4\le12n^3+28n^2+20n+4$). Base $u_1=\frac12\le\frac12$.