Question

Find the value of $ {(1.01)^5} $ correct to $ 5 $ decimal places.

Answer 1

Hint: For this type of problem we use a binomial expansion formula. For this we first write the given problem in terms of $ {(x + a)^n} $ and then on expanding and simplifying to get the required solution of the problem.
\[{(x + a)^n} = {\,^n}{C_0}{x^n} + {\,^n}{C_1}{x^{n - 1}}{a^1} + {\,^n}{C_2}{x^{n - 2}}{a^2} + ..... + {\,^n}{C_n}{a^n}\]where ‘x’ is the first and ‘a’ is the second part of the base.
Formulas of combination: $ ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $

Complete step-by-step answer:
Given, $ {(1.01)^5} $
First we convert base in terms of (x + a) to get the value of ‘x’ and ‘a’.
Writing $ 1.01\,\,as\,\,\left( {1 + 0.01} \right) $
Therefore, $ {(1.01)^5} $ becomes $ {\left( {1 + 0.01} \right)^5} $ .
Then, according to binomial $ {(x + a)^n} $ . We have $ x = 1\,\,and\,\,a = 0.01 $ .
Now, substituting values of ‘x’ and ‘a’ in above mentioned binomial formula. We have,
 $ {\left( {1 + 0.01} \right)^5}{ = ^5}{C_0}{(1)^5}{ + ^5}{C_1}{(1)^4}{\left( {0.01} \right)^1}{ + ^5}{C_2}{(1)^3}{\left( {0.01} \right)^2}{ + ^5}{C_3}{(1)^2}{\left( {0.01} \right)^3}{ + ^5}{C_4}{(1)^1}{(0.01)^4}{ + ^5}{C_5}{(0.01)^5} $
Simplifying the right hand side of the above formed equation by using values of combination.
 $ ^5{C_0}{ = ^5}{C_5} = 1,\,\,{\,^5}{C_1}{ = ^5}{C_4} = 5,\,\,\,and\,{\,^5}{C_2}{ = ^5}{C_3} = \dfrac{{5!}}{{2!3!}} = 10 $
Using these values of combination in above formed equation. We have,
 $ {\left( {1 + 0.1} \right)^5} = (1)(1) + (5)(0.01) + (10)(0.0001) + (10)(0.000001) + (5)(0.00000001) + (1)(0.0000000001) $
Simplifying the right hand side of the above formed equation.
 $ {\left( {1 + 0.01} \right)^5} = 1 + 0.05 + 0.001 + 0.00001 + 0.00000005 + 0.0000000001 $
 $ \Rightarrow {\left( {1 + 0.1} \right)^5} = $ $ 1.50101001501 $
Or
 $ {\left( {1.01} \right)^5} = 1.05101001501 $
Hence, from above we see that the value of $ {\left( {1.01} \right)^5} $ is $ 1.05101001501 $ .
But, its value up to $ 5 $ decimals places is $ = 1.05101 $

Note: In binomial there are two expansion formulas. One for those terms in which index power or binomial power is a natural number, for this binomial expansion formula is\[{(x + a)^n} = {\,^n}{C_0}{x^n} + {\,^n}{C_1}{x^{n - 1}}{a^1} + {\,^n}{C_2}{x^{n - 2}}{a^2} + ..... + {\,^n}{C_n}{a^n}\]. But, if index power or binomial power is either negative or in fraction. Then binomial expansion will be given as:
\[{\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n.(n - 1)}}{{2!}}{x^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{x^3} + ..... + \dfrac{{n(n - 1)(n - 2)...(n - r + 1)}}{{r!}}{x^r} + ...\].
In this expansion the number of terms are infinite. So, students must choose appropriate formulas of expansion to find the correct solution of a problem.