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

Find the value of (1.01)5{(1.01)^5} correct to 55 decimal places.

Explanation

Solution

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{(x + a)^n} and then on expanding and simplifying to get the required solution of the problem.
(x+a)n= nC0xn+ nC1xn−1a1+ nC2xn−2a2+.....+ nCnan{(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: nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}

Complete step-by-step answer :
Given, (1.01)5{(1.01)^5}
First we convert base in terms of (x + a) to get the value of ‘x’ and ‘a’.
Writing 1.01  as  (1+0.01)1.01\,\,as\,\,\left( {1 + 0.01} \right)
Therefore, (1.01)5{(1.01)^5} becomes (1+0.01)5{\left( {1 + 0.01} \right)^5} .
Then, according to binomial (x+a)n{(x + a)^n} . We have x=1  and  a=0.01x = 1\,\,and\,\,a = 0.01 .
Now, substituting values of ‘x’ and ‘a’ in above mentioned binomial formula. We have,
(1+0.01)5=5C0(1)5+5C1(1)4(0.01)1+5C2(1)3(0.01)2+5C3(1)2(0.01)3+5C4(1)1(0.01)4+5C5(0.01)5{\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.
5C0=5C5=1,   5C1=5C4=5,   and  5C2=5C3=5!2!3!=10^5{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,
(1+0.1)5=(1)(1)+(5)(0.01)+(10)(0.0001)+(10)(0.000001)+(5)(0.00000001)+(1)(0.0000000001){\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.
(1+0.01)5=1+0.05+0.001+0.00001+0.00000005+0.0000000001{\left( {1 + 0.01} \right)^5} = 1 + 0.05 + 0.001 + 0.00001 + 0.00000005 + 0.0000000001
⇒(1+0.1)5=\Rightarrow {\left( {1 + 0.1} \right)^5} = 1.501010015011.50101001501
Or
(1.01)5=1.05101001501{\left( {1.01} \right)^5} = 1.05101001501
Hence, from above we see that the value of (1.01)5{\left( {1.01} \right)^5} is 1.051010015011.05101001501 .
But, its value up to 55 decimals places is =1.05101= 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= nC0xn+ nC1xn−1a1+ nC2xn−2a2+.....+ nCnan{(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:
(1+x)n=1+nx+n.(n−1)2!x2+n(n−1)(n−2)3!x3+.....+n(n−1)(n−2)...(n−r+1)r!xr+...{\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.