Question

Using the Binomial Theorem, evaluate ${{\left( 101 \right)}^{4}}$

Answer 1

In the question we will first separate the numbers (in form of sum) and keep one of the numbers as $1$ . Now the formula we will use to find the value of the question is:
${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{1}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n}}{{b}^{n}}$
where $a$ and $b$ are the sum of the value of the numbers given in the question and one of them is $1$ for easier calculation while $n$ is the power upto which the binomial is evaluated.

Complete step-by-step answer:
Simplifying the values of the binomial theorem by placing the value of $n=4$
${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{1}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n}}{{b}^{n}}$
${{\left( a+b \right)}^{n}}={}^{4}{{C}_{0}}{{a}^{4}}+{}^{4}{{C}_{1}}{{a}^{4-1}}{{b}^{1}}+{}^{4}{{C}_{2}}{{a}^{4-2}}{{b}^{2}}+{}^{4}{{C}_{3}}{{a}^{4-3}}{{b}^{3}}+{}^{4}{{C}_{4}}{{a}^{4-4}}{{b}^{4}}$
Placing the values of $a,b$ in the formula as $(a+b)=(100+1)$ respectively, we get:
$\Rightarrow {}^{4}{{C}_{0}}{{a}^{4}}+{}^{4}{{C}_{1}}{{a}^{3}}{{b}^{1}}+{}^{4}{{C}_{2}}{{a}^{2}}{{b}^{2}}+{}^{4}{{C}_{3}}{{a}^{1}}{{b}^{3}}+{}^{4}{{C}_{4}}{{a}^{0}}{{b}^{4}}$
${{\left( 100+1 \right)}^{n}}={}^{4}{{C}_{0}}{{\left( 100 \right)}^{4}}+{}^{4}{{C}_{1}}{{\left( 100 \right)}^{3}}{{1}^{1}}+{}^{4}{{C}_{2}}{{\left( 100 \right)}^{2}}{{1}^{2}}+{}^{4}{{C}_{3}}{{\left( 100 \right)}^{1}}{{1}^{3}}+{}^{4}{{C}_{4}}{{\left( 100 \right)}^{0}}{{1}^{4}}$
Calculating the R.H.S of the binomial theorem and solving the combination of all the five combinations ${}^{4}{{C}_{0}},{}^{4}{{C}_{1}},{}^{4}{{C}_{2}},{}^{4}{{C}_{3}},{}^{4}{{C}_{4}}$ we get the value as:
$\Rightarrow 1{{\left( 100 \right)}^{4}}+\frac{4}{1}{{\left( 100 \right)}^{3}}{{1}^{1}}+\frac{4.3}{2.1}{{\left( 100 \right)}^{2}}{{1}^{2}}+\frac{4.3.2}{3.2.1}{{\left( 100 \right)}^{1}}{{1}^{3}}+{{\left( 100 \right)}^{0}}{{1}^{4}}$
$\Rightarrow {{\left( 100 \right)}^{4}}+4{{\left( 100 \right)}^{3}}+6{{\left( 100 \right)}^{2}}+4{{\left( 100 \right)}^{1}}+{{\left( 100 \right)}^{0}}$
$\Rightarrow 100000000+4000000+60000+400+1$
$\Rightarrow 104060401$

Hence, the value of ${{\left( 101 \right)}^{4}}$ using binomial theorem is $104060401$.

Note: Students may go wrong while solving the powers and the combination, remember the first part of the combination will always be ${{a}^{n}}$ and last part be ${{b}^{n}}$ and the power of $a$ will decrease and power of $b$ will increase.