Question

Using binomial theorem, evaluate \[{{\left( 102 \right)}^{5}}\].

Answer 1

Hint: In this problem, we have to evaluate the given number with the fifth root. We can first split the given term in the form of \[{{\left( a+b \right)}^{n}}\] We know that the binomial expansion of \[{{\left( a+b \right)}^{n}}\] is \[{{\left( a+b \right)}^{n}}{{=}^{n}}{{C}_{r}}{{a}^{n}}{{b}^{0}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......{{+}^{n}}{{C}_{n}}{{a}^{0}}{{b}^{n}}\]. Here n is the power term. We know that \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] , we can find all the combinations and substitute the value of combination, a and b value to get the answer.

Complete step-by-step answer:
Here we have to evaluate\[{{\left( 102 \right)}^{5}}\] using a binomial theorem.
We can now write the give number in the form \[{{\left( a+b \right)}^{n}}\], we get
\[{{\left( 100+2 \right)}^{5}}\], where a = 100, b = 2 and n = 5.
We know that the binomial expansion is
\[{{\left( a+b \right)}^{n}}{{=}^{n}}{{C}_{r}}{{a}^{n}}{{b}^{0}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......{{+}^{n}}{{C}_{n}}{{a}^{0}}{{b}^{n}}\]
We can now write this expansion with n = 5, we get
\[{{\left( a+b \right)}^{5}}{{=}^{5}}{{C}_{0}}{{a}^{5}}{{b}^{0}}{{+}^{5}}{{C}_{1}}{{a}^{4}}{{b}^{1}}{{+}^{5}}{{C}_{2}}{{a}^{3}}{{b}^{2}}{{+}^{5}}{{C}_{3}}{{a}^{2}}{{b}^{3}}{{+}^{5}}{{C}_{4}}{{a}^{1}}{{b}^{4}}{{+}^{5}}{{C}_{5}}{{a}^{0}}{{b}^{5}}\]…….. (1)
We can find the combination value using the formula \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\], where n = 5 and r = 0
\[{{\Rightarrow }^{5}}{{C}_{0}}=\dfrac{5!}{0!\left( 5-0 \right)!}=\dfrac{5!}{5!}=1\]
When n = 5 and r =1, we get
\[{{\Rightarrow }^{5}}{{C}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!}=\dfrac{5!}{4!}=5\]
Similarly, for r = 2, 3, 4, 5, we will get the combinations value as
\[\begin{align}
  & {{\Rightarrow }^{5}}{{C}_{2}}=\dfrac{5!}{2!\left( 5-2 \right)!}=\dfrac{5!}{2!3!}=10 \\
 & {{\Rightarrow }^{5}}{{C}_{3}}=\dfrac{5!}{3!\left( 5-3 \right)!}=\dfrac{5!}{3!2!}=10 \\
 & {{\Rightarrow }^{5}}{{C}_{4}}=\dfrac{5!}{4!\left( 5-4 \right)!}=\dfrac{5!}{4!}=5 \\
 & {{\Rightarrow }^{5}}{{C}_{5}}=\dfrac{5!}{5!\left( 5-5 \right)!}=\dfrac{5!}{5!}=1 \\
\end{align}\]
We can now substitute the value of a, b, n and the combinations value in (1), we get
\[\Rightarrow {{\left( 100+2 \right)}^{5}}=1{{\left( 100 \right)}^{5}}{{\left( 2 \right)}^{0}}+5{{\left( 100 \right)}^{4}}{{\left( 2 \right)}^{1}}+10{{\left( 100 \right)}^{3}}{{\left( 2 \right)}^{2}}+10{{\left( 100 \right)}^{2}}{{\left( 2 \right)}^{3}}+5{{\left( 100 \right)}^{1}}{{\left( 2 \right)}^{4}}+1{{\left( 2 \right)}^{5}}\]
We can now simplify the above step, we get
\[\Rightarrow {{\left( 100+2 \right)}^{5}}=10000000000+1000000000+40000000+800000+8000+32\]
We can now add the above terms, we get
\[\Rightarrow {{\left( 102 \right)}^{5}}=11040808032\]
Therefore, the value of \[{{\left( 102 \right)}^{5}}=11040808032\]

Note: We should always remember that for large number of digits, the power of the given term can be easily found by using the binomial theorem expansion 0\[{{\left( a+b \right)}^{n}}{{=}^{n}}{{C}_{r}}{{a}^{n}}{{b}^{0}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......{{+}^{n}}{{C}_{n}}{{a}^{0}}{{b}^{n}}\] and we should know to find the value of the combination part using the formula \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].