Question

How do you use binomial theorem to approximate ${\left( {1.02} \right)^8}?$

Answer 1

Hint: First split the given number (base) into whole and decimal part or fraction and then use the binomial expansion in order to expand it. Calculate the values with respective values of binomial coefficients and finally you will get the desired result.Binomial expansion of ${\left( {1 + x} \right)^n}$ is given as following
${\left( {1 + x} \right)^n} = \sum\limits_{r = 0}^n {\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right){x^r}} ,\;{\text{where}}\;\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right)$ is called the binomial coefficient and calculated as $\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right) = \dfrac{{n!}}{{r!(n - r)!}}$

Complete step by step answer:
To find the approximate value of ${\left( {1.02} \right)^8}$ with help of binomial theorem, we have to first split the base ${\text{(i}}{\text{.e}}{\text{.}}\;{\text{1}}{\text{.02)}}$ into sum of whole number and decimal number.We can see that the number $1.02$ is a sum of $1\;{\text{and}}\;0.2$ so we can write it as follows,
$1.02 = 1 + 0.02$ we will write $0.02$ in fractional form to make the calculations easy
$0.02 = \dfrac{2}{{100}} = \dfrac{1}{{50}}$
Therefore $1.02$ will be written as
$
1.02 = 1 + 0.02 = 1 + \dfrac{1}{{50}} \\
\Rightarrow {\left( {1.02} \right)^8} = {\left( {1 + \dfrac{1}{{50}}} \right)^8} \\
 $
With the help of binomial theorem we can expand,
${\left( {1 + x} \right)^n} = \sum\limits_{r = 0}^n {\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right){x^r}} ,\;{\text{where}}\;\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right)$
$\Rightarrow {\left( {1 + \dfrac{1}{{50}}} \right)^8} = \sum\limits_{r = 0}^8 {\left( {\begin{array}{*{20}{c}}
  8 \\
  r
\end{array}} \right){{\left( {\dfrac{1}{{50}}} \right)}^r}} $
Evaluating the summation function we will get,
$
\Rightarrow\sum\limits_{r = 0}^8 {\left( {\begin{array}{*{20}{c}}
  8 \\
  r
\end{array}} \right){{\left( {\dfrac{1}{{50}}} \right)}^r}} \\
\Rightarrow\left( {\begin{array}{*{20}{c}}
  8 \\
  0
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^0} + \left( {\begin{array}{*{20}{c}}
  8 \\
  1
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^1} + \left( {\begin{array}{*{20}{c}}
  8 \\
  2
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^2} + \left( {\begin{array}{*{20}{c}}
  8 \\
  3
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^3} + \left( {\begin{array}{*{20}{c}}
  8 \\
  4
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^4} + \left( {\begin{array}{*{20}{c}}
  8 \\
  5
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^5} + \\
\left( {\begin{array}{*{20}{c}}
  8 \\
  6
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^6} + \left( {\begin{array}{*{20}{c}}
  8 \\
  7
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^7} + \left( {\begin{array}{*{20}{c}}
  8 \\
  8
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^8} \\
 $
We can calculate value of a binomial coefficient as,
$\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right) = \dfrac{{n!}}{{r!(n - r)!}}$
Therefore calculating values for above binomial coefficients
$
\left( {\begin{array}{*{20}{c}}
  8 \\
  0
\end{array}} \right) = \dfrac{{8!}}{{0!(8 - 0)!}} = \dfrac{{8!}}{{1 \times 8!}} = 1 \\
\Rightarrow\left( {\begin{array}{*{20}{c}}
  8 \\
  1
\end{array}} \right) = \dfrac{{8!}}{{1!(8 - 1)!}} = \dfrac{{8!}}{{1 \times 7!}} = 8 \\
\Rightarrow \left( {\begin{array}{*{20}{c}}
  8 \\
  2
\end{array}} \right) = \dfrac{{8!}}{{2!(8 - 2)!}} = \dfrac{{8!}}{{2 \times 6!}} = 28 \\
\Rightarrow \left( {\begin{array}{*{20}{c}}
  8 \\
  3
\end{array}} \right) = \dfrac{{8!}}{{3!(8 - 3)!}} = \dfrac{{8!}}{{6 \times 5!}} = 56 \\
\Rightarrow\left( {\begin{array}{*{20}{c}}
  8 \\
  4
\end{array}} \right) = \dfrac{{8!}}{{4!(8 - 4)!}} = \dfrac{{8!}}{{24 \times 4!}} = 70 \\
\Rightarrow \left( {\begin{array}{*{20}{c}}
  8 \\
  5
\end{array}} \right) = \dfrac{{8!}}{{5!(8 - 5)!}} = \dfrac{{8!}}{{120 \times 3!}} = 56 \\
\Rightarrow\left( {\begin{array}{*{20}{c}}
  8 \\
  6
\end{array}} \right) = \dfrac{{8!}}{{6!(8 - 6)!}} = \dfrac{{8!}}{{720 \times 2!}} = 28 \\
\Rightarrow\left( {\begin{array}{*{20}{c}}
  8 \\
  7
\end{array}} \right) = \dfrac{{8!}}{{7!(8 - 7)!}} = \dfrac{{8!}}{{5040 \times 1!}} = 8 \\
\Rightarrow\left( {\begin{array}{*{20}{c}}
  8 \\
  8
\end{array}} \right) = \dfrac{{8!}}{{8!(8 - 8)!}} = \dfrac{{8!}}{{40320 \times 0!}} = 1 \\
 $
Now substituting the values of binomial coefficients in the above binomial expansion
$
\left( {\begin{array}{*{20}{c}}
  8 \\
  0
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^0} + \left( {\begin{array}{*{20}{c}}
  8 \\
  1
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^1} + \left( {\begin{array}{*{20}{c}}
  8 \\
  2
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^2} + \left( {\begin{array}{*{20}{c}}
  8 \\
  3
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^3} + \left( {\begin{array}{*{20}{c}}
  8 \\
  4
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^4} + \left( {\begin{array}{*{20}{c}}
  8 \\
  5
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^5} + \\
\left( {\begin{array}{*{20}{c}}
  8 \\
  6
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^6} + \left( {\begin{array}{*{20}{c}}
  8 \\
  7
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^7} + \left( {\begin{array}{*{20}{c}}
  8 \\
  8
\end{array}} \right){\left( {\dfrac{1}{{50}}} \right)^8} \\
\Rightarrow 1{\left( {\dfrac{1}{{50}}} \right)^0} + 8{\left( {\dfrac{1}{{50}}} \right)^1} + 28{\left( {\dfrac{1}{{50}}} \right)^2} + 56{\left( {\dfrac{1}{{50}}} \right)^3} + 70{\left( {\dfrac{1}{{50}}} \right)^4} + 56{\left( {\dfrac{1}{{50}}} \right)^5} + \\
  28{\left( {\dfrac{1}{{50}}} \right)^6} + 8{\left( {\dfrac{1}{{50}}} \right)^7} + 1{\left( {\dfrac{1}{{50}}} \right)^8} \\
\Rightarrow 1 + \dfrac{8}{{50}} + \dfrac{{28}}{{{{50}^2}}} + \dfrac{{56}}{{{{50}^3}}} + \dfrac{{70}}{{{{50}^4}}} + \dfrac{{56}}{{{{50}^5}}} + \dfrac{{28}}{{{{50}^6}}} + \dfrac{8}{{{{50}^7}}} + \dfrac{1}{{{{50}^8}}} \\
\therefore\dfrac{{45767944570401}}{{39062600000000}} \approx 1.17166 \\
 $
The final value is written on an approximation of up to five decimal places.
You can check the answer by calculating the given exponential term ${\left( {1.02} \right)^8}$ with help of an advanced calculator.

Note:As the power increases the expansion becomes lengthy and tedious to calculate. A binomial expression that has been raised to a very large power can be easily calculated with the help of Binomial Theorem.Binomial coefficients $\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right)$ are sometimes also written as ${}^n{C_r}$ and also we can directly find the values of a binomial coefficient $\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right)$.