Find whether ${{\left( 100 \right)}^{50}}+{{\left( 99 \right)}^{50}}$ is:
(a) $<{{\left( 101 \right)}^{50}}$
(b) $<\left( 101 \right)$
(c) $>{{\left( 101 \right)}^{50}}$
(d) $>\left( 101 \right)$

Hint: Use binomial expansion to expand ${{\left( 101 \right)}^{50}}$ and ${{\left( 99 \right)}^{50}}$ . Then subtract the two expressions obtained to get a new expression. Use the elements of the new expression to arrive at a relation between ${{\left( 100 \right)}^{50}}+{{\left( 99 \right)}^{50}}$ and ${{\left( 101 \right)}^{50}}$ .

Complete step-by-step answer:
We need to find the relation between ${{\left( 100 \right)}^{50}}+{{\left( 99 \right)}^{50}}$ and ${{\left( 101 \right)}^{50}}$ .
We will use binomial expansion to solve this question.
First, let us define binomial expansion.
The binomial expansion (or binomial theorem) describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand ${{\left( x+y \right)}^{n}}$ into a sum involving terms of the form $a{{x}^{b}}{{y}^{c}}$ , where the exponents b and c are non negative integers with b + c = n, and the coefficient a of each term is a specific positive integer depending on n and b.
Binomial expansion gives us the formula:
${{\left( a+b \right)}^{n}}={}_{0}^{n}C\cdot {{a}^{n}}{{b}^{0}}+{}_{1}^{n}C\cdot {{a}^{n-1}}{{b}^{1}}+...+{}_{n}^{n}C\cdot {{a}^{0}}{{b}^{n}}$
 Let us first evaluate ${{\left( 101 \right)}^{50}}$
${{\left( 101 \right)}^{50}}={{\left( 100+1 \right)}^{50}}$
${{\left( 101 \right)}^{50}}={{100}^{50}}+{}_{1}^{50}C\cdot {{100}^{49}}+{}_{2}^{50}C\cdot {{100}^{48}}+...+1$$\begin{align}
  & {{\left( 101 \right)}^{50}}={{\left( 100+1 \right)}^{50}} \\
 & {{\left( 101 \right)}^{50}}={{100}^{50}}+{}_{1}^{50}C\cdot {{100}^{49}}+{}_{2}^{50}C\cdot {{100}^{48}}+...+1 \\
 & \\
\end{align}$ …(1)
Now, we will evaluate ${{\left( 99 \right)}^{50}}$
${{\left( 99 \right)}^{50}}={{\left( 100-1 \right)}^{50}}$
${{\left( 99 \right)}^{50}}={{100}^{50}}-{}_{1}^{50}C\cdot {{100}^{49}}+{}_{2}^{50}C\cdot {{100}^{48}}-...+1$ …(2)
We will now subtract equation (2) from equation (1) .
${{\left( 101 \right)}^{50}}-{{\left( 99 \right)}^{50}}=\left( {{100}^{50}}+{}_{1}^{50}C\cdot {{100}^{49}}+{}_{2}^{50}C\cdot {{100}^{48}}+...+1 \right)-\left( {{100}^{50}}-{}_{1}^{50}C\cdot {{100}^{49}}+{}_{2}^{50}C\cdot {{100}^{48}}-...+1 \right)$${{\left( 101 \right)}^{50}}-{{\left( 99 \right)}^{50}}=\left( {{100}^{50}}+{}_{1}^{50}C\cdot {{100}^{49}}+{}_{2}^{50}C\cdot {{100}^{48}}+...+1 \right)-{{100}^{50}}+{}_{1}^{50}C\cdot {{100}^{49}}-{}_{2}^{50}C\cdot {{100}^{48}}+...-1$${{\left( 101 \right)}^{50}}-{{\left( 99 \right)}^{50}}=2\times \left( {}_{1}^{50}C\cdot {{100}^{49}}+{}_{3}^{50}C\cdot {{100}^{47}}+... \right)$
${{\left( 101 \right)}^{50}}-{{\left( 99 \right)}^{50}}=2\times {}_{1}^{50}C\cdot {{100}^{49}}+2\times \left( {}_{3}^{50}C\cdot {{100}^{47}}+{}_{5}^{50}C\cdot {{100}^{45}}+... \right)$
As we know that ${}_{1}^{50}C=50$ . Substituting this in above equation, we will get the following:
${{\left( 101 \right)}^{50}}-{{\left( 99 \right)}^{50}}=2\times 50\times {{100}^{49}}+2\times \left( {}_{3}^{50}C\cdot {{100}^{47}}+{}_{5}^{50}C\cdot {{100}^{45}}+... \right)$
As $2\times 50=100$ . Substituting this in above equation, we will get the following:
${{\left( 101 \right)}^{50}}-{{\left( 99 \right)}^{50}}=100\times {{100}^{49}}+2\times \left( {}_{3}^{50}C\cdot {{100}^{47}}+{}_{5}^{50}C\cdot {{100}^{45}}+... \right)$
${{\left( 101 \right)}^{50}}-{{\left( 99 \right)}^{50}}={{100}^{50}}+2\times \left( {}_{3}^{50}C\cdot {{100}^{47}}+{}_{5}^{50}C\cdot {{100}^{45}}+... \right)$
Now, let $P=2\times \left( {}_{3}^{50}C\cdot {{100}^{47}}+{}_{5}^{50}C\cdot {{100}^{45}}+... \right)$ , where P is some positive number.
Substituting this in above equation, we will get the following:
 ${{\left( 101 \right)}^{50}}-{{\left( 99 \right)}^{50}}={{100}^{50}}+P$
Also, we know that:
${{100}^{50}}+P>{{100}^{50}}$
Using this condition in the above equation, we will get the following:
${{\left( 101 \right)}^{50}}-{{\left( 99 \right)}^{50}}>{{\left( 100 \right)}^{50}}$
Or, ${{\left( 101 \right)}^{50}}>{{\left( 100 \right)}^{50}}+{{\left( 99 \right)}^{50}}$
Or, ${{\left( 100 \right)}^{50}}+{{\left( 99 \right)}^{50}}<{{\left( 101 \right)}^{50}}$ .
Hence, option (a) is the correct answer.

Note: It is important to understand the fact that alternate terms in the expansions of ${{\left( 101 \right)}^{50}}$ and ${{\left( 99 \right)}^{50}}$ will be of opposite sign. So, if we subtract these expansions, these alternate terms will get cancelled out. Use this property to arrive at the final answer.