Question

Find the value of \[{\left( {1.03} \right)^{\dfrac{1}{3}}}\] up to four decimal places.

Answer 1

Hint: Split the given expression, then apply the binomial theorem.
In algebra, the binomial theorem describes the method of expanding expression which has been raised by finite power. The numbers in each row in the pascal triangle are known as the binomial coefficients.\[(1.03)\] Can be expressed as the sum or difference of two numbers and the binomial theorem can be applied. The given question can be written as \[(1.03)\]\[ = \]\[(1 + 0.03)\].

FORMULA TO BE USED:
Use the formula
\[{\left( {1 + x} \right)^n} = 1 + nx + n\left( {\dfrac{{n - 1}}{{2!}}} \right){x^1} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^2} + .......{x^4}\]

Complete step-by-step answer:
We can split the given expression as
\[\left( {1.03} \right) = \left( {1 + 0.03} \right)\]
Now to find the value of \[{\left( {1.03} \right)^{\dfrac{1}{3}}}\] we will apply the binomial formula in the given expression
Which is,
\[{\left( {1 + x} \right)^n} = 1 + nx + n\left( {\dfrac{{n - 1}}{{2!}}} \right){x^1} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^2} + .......{x^4}\]
Putting required value in given binomial formula,
=\[{\left( {1 + 0.03} \right)^{\dfrac{1}{3}}} = 1 + \dfrac{1}{3} \times 0.03 + \dfrac{1}{3} \times \left( {\dfrac{1}{3} - 1} \right) \times {\left( {0.03} \right)^2} \times \dfrac{1}{{2!}} + \left( {\dfrac{1}{3} - 1} \right)\left( {\dfrac{1}{3} - 2} \right) \times {\left( {0.03} \right)^2} \times \dfrac{1}{{3!}} + .........]\]
We are only required to solve 4 terms as we are asked to find 4 decimal places. no need to solve all.
Now we know that,
\[2! = 2 \times 1\]
\[3! = 3 \times 2 \times 1\]
Therefore, the value of 2! is 2, and 3! is 6
=\[1 + 0.01 + \dfrac{1}{3} \times \dfrac{{ - 2}}{3} \times \dfrac{{0.00090}}{2} + \dfrac{1}{3} \times \dfrac{{ - 2}}{3} \times \dfrac{{ - 5}}{3} \times \dfrac{{0.0003}}{6} + .......\]
Further simplifying more, we get,
=\[1 + 0.0100 - 0.00010 + .......\]
=\[1.00990\]
=1.0099 (up to 4 decimal places)
Therefore, the value \[{\left( {1.03} \right)^{\dfrac{1}{3}}}\] is 1.0099 up to four decimal places.

Note: Always try to split the expression which can be suitable for the binomial theorem. In this type of question, we always split the expression in the form of the addition of two numbers to make our steps easy. In the binomial expression, don’t calculate the whole series. To find a binomial theorem you can also use Pascal's triangle. In Pascal's triangle, every row is built above another, it gives us the coefficient in \[{\left( {a + b} \right)^n}\] form. We can use these coefficients to find the entire expanded expression.