Question

If $a={2021}^{1010}$ and $b={2019}^{1010}+{2020}^{1010}$, then which of the following is correct?

• A. $a<b$
• B. $a=b$
• C. $a>b$
• D. $a-b$ is an odd integer.

Answer 1

The correct option is C $a>b$

Given : $a={2021}^{1010},b={2019}^{1010}+{2020}^{1010}$
Now,
${2021}^{1010}-{2019}^{1010}=(2020+1{)}^{1010}-(2020-1{)}^{1010}=2[T_2+T_4+T_6+\cdots ]=2\left[{}^{1010}{C}_1\right(2020{)}^{1009}+{}^{1010}{C}_3(2020{)}^{1007}+\cdots ]=(2020{)}^{1010}+2\left[{}^{1010}{C}_3\right(2020{)}^{1007}+\cdots ]$
Clearly, $a-b$ is an even integer.
Therefore,
${2021}^{1010}-{2019}^{1010}>{2020}^{1010}\Rightarrow {2021}^{1010}>{2019}^{1010}+{2020}^{1010}\therefore a>b$