Question

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

Answer 1

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

& {{\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{aligned}$$ We can now substitute the value of a, b, n and the combinations value in (1), we get $$\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)!}$$.