Question

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

Answer 1

Putting $(101{)}^{50}$=a and $({100}^{50}+{99}^{50})=b$, we get

$(a-b)=(101{)}^{50}-(100{)}^{50}-(99{)}^{50}$

$=(101{)}^{50}-(99{)}^{50}-(100{)}^{50}=(100+1{)}^{50}-(100-1{)}^{50}-(100{)}^{50}$

$=2\times {[}^{50}{C}_1\times {100}^{49}{+}^{50}{C}_3\times {100}^{47}+....{+}^{50}{C}_{49}\times 100]-(100{)}^{50}$

$=[2{\times }^{50}{C}_3\times {100}^{47}+2{\times }^{50}{C}_5\times {100}^{45}+...+2{\times }^{50}{C}_{49}\times 100]$

=(a positive integer).

Thus, $\Rightarrow a>b\Rightarrow (101{)}^{50}>({100}^{50}+{99}^{50})$.