Question

Using binomial theorem, prove that $(101{)}^{50}>{100}^{50}+{99}^{50}$.

Answer 1

We have,

$x={\left(101\right)}^{50}>{100}^{50}+{99}^{50}$

Let $x>y$

So, $x={101}^{50},y={100}^{50}+{99}^{50}$

Therefore,

$x-y={\left(101\right)}^{50}-{\left(100\right)}^{50}-{\left(99\right)}^{50}={\left(100+1\right)}^{50}-{\left(100-1\right)}^{50}-{100}^{50}=\left({}^{50}{C}_{1‘}{100}^{50}{1}^0+{}^{50}{C}_1{100}^{49}{\left(1\right)}^1+{}^{50}{C}_2{100}^{48}{1}^2.......\right)-\left({}^{50}{C}_0{100}^{50}-{}^{50}{C}_1{100}^{49}+{}^{50}{C}_2{100}^{48}\right)-{100}^{50}=2\left({}^{50}{C}_1{100}^{49}{+}^{50}{C}_3{100}^{47}{\left(1\right)}^3+{}^{50}{C}_5{100}^{45}.....\right)-{100}^{50}={100}^{50}+2\left({}^{50}{C}_3{100}^{47}+{}^{50}{C}_5{100}^{45}.......\right)-{100}^{50}$

$x-y=2\left({}^{50}{C}_3{100}^{47}+{}^{50}{C}_5{100}^{45}........\right)$

Therefore,

$x-y=a+ve$ number

$x-y>0,{101}^{50}-\left({100}^{50}+{99}^{50}\right)>0\Rightarrow {101}^{50}>{100}^{50}+{99}^{50}$

Hence, proved.