Question

The larger of ${(2n+1)}^n$ and ${(2n-1)}^n+{2n}^n$ is (Assume that $n$ is a positive integer greater than 100).

• A. ${(2n+1)}^n$
• B. ${(2n-1)}^n+{2n}^n$
• C. Both are equal
• D. None of these

Answer 1

The correct option is A ${(2n+1)}^n$

Given $(2n+1{)}^n$ and $(2n-1{)}^n+(2n{)}^n$
Let $n=200$
So, $(401{)}^{200}$ and $(399{)}^{200}+(400{)}^{200}$
Consider, $(401{)}^{200}-(399{)}^{200}$
$=(400+1{)}^{200}-(400-1{)}^{200}$
$={[}^{200}{C}_0{+}^{200}{C}_1\left(400{)}^{199}{+}^{200}{C}_2\right(400{)}^{198}+.....+1]-{[}^{200}{C}_0{-}^{200}{C}_1(400{)}^{199}{+}^{200}{C}_2(400{)}^{198}+.....+1]$
$=2{[}^{200}{C}_1\left(400{)}^{199}{+}^{200}{C}_3\right(400{)}^{197}+....{+}^{200}{C}_{199}\left(400\right)]$
$=2.200(400{)}^{199}+2.200{[}^{200}{C}_3(400{)}^{196}+....{+}^{200}{C}_{198}\left(400\right)]$
$=(400{)}^{200}+\text{a positive number}>(400{)}^{200}$
$\Rightarrow (401{)}^{200}-(399{)}^{200}>(400{)}^{200}$
$\Rightarrow (401{)}^{200}>(399{)}^{200}+(400{)}^{200}$
Hence, $(2n+1{)}^n>(2n-1{)}^n+(2n{)}^n$