Question

How do you use the binomial theorem to approximate ${\left( {2.99} \right)^{12}}$?

Answer 1

Hint: For solving this question, first we need to understand the binomial theorem. But before that we will convert the given term into the binomial form which is ${\left( {a + b} \right)^n}$. After that we will apply the formula given by the theorem and find our answer.

Formula used:
Binomial theorem states that for any positive integer $n$, the ${n^{th}}$ power of the sum of two numbers $a$ and $b$ may be expressed as the sum of $n + 1$ terms of the form:
$\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right){a^{n - r}}{b^r}$
In this sequence of terms, the index r takes on the successive values \[0,1,2, \ldots ,n\]. The coefficients, called the binomial coefficients, are defined by the formula:
$\left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right) = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$

Complete step by step solution:
We know that we can use the binomial theorem only when the given term is in the binomial form.
Therefor we will first convert the given term into the binomial form.
We can write ${\left( {2.99} \right)^{12}} = {\left( {3 - 0.01} \right)^{12}}$.
Thus, here we have $a = 3$, $b = - 0.01$ and $n = 12$.
Now, we will apply the formula ${\left( {a + b} \right)^n} = \left( {\begin{array}{*{20}{c}}
  n \\
  r
\end{array}} \right){a^{n - r}}{b^r}$.
\[
   \Rightarrow {\left( {3 - 0.01} \right)^{12}} \\
   = \left( {\begin{array}{*{20}{c}}
  {12} \\
  0
\end{array}} \right)\left( {{3^{12}}} \right)\left( { - {{0.01}^0}} \right) + \left( {\begin{array}{*{20}{c}}
  {12} \\
  1
\end{array}} \right)\left( {{3^{11}}} \right)\left( { - {{0.01}^1}} \right) + \left( {\begin{array}{*{20}{c}}
  {12} \\
  2
\end{array}} \right)\left( {{3^{10}}} \right)\left( { - {{0.01}^2}} \right) + \left( {\begin{array}{*{20}{c}}
  {12} \\
  3
\end{array}} \right)\left( {{3^9}} \right)\left( { - {{0.01}^3}} \right) \\
   + \left( {\begin{array}{*{20}{c}}
  {12} \\
  4
\end{array}} \right)\left( {{3^8}} \right)\left( { - {{0.01}^4}} \right) + \left( {\begin{array}{*{20}{c}}
  {12} \\
  5
\end{array}} \right)\left( {{3^7}} \right)\left( { - {{0.01}^5}} \right) + \left( {\begin{array}{*{20}{c}}
  {12} \\
  6
\end{array}} \right)\left( {{3^6}} \right)\left( { - {{0.01}^6}} \right) + \left( {\begin{array}{*{20}{c}}
  {12} \\
  7
\end{array}} \right)\left( {{3^5}} \right)\left( { - {{0.01}^7}} \right) \\
   + \left( {\begin{array}{*{20}{c}}
  {12} \\
  8
\end{array}} \right)\left( {{3^4}} \right)\left( { - {{0.01}^8}} \right) + \left( {\begin{array}{*{20}{c}}
  {12} \\
  9
\end{array}} \right)\left( {{3^3}} \right)\left( { - {{0.01}^9}} \right) + \left( {\begin{array}{*{20}{c}}
  {12} \\
  {10}
\end{array}} \right)\left( {{3^2}} \right)\left( { - {{0.01}^{10}}} \right) + \left( {\begin{array}{*{20}{c}}
  {12} \\
  {11}
\end{array}} \right)\left( {{3^1}} \right)\left( { - {{0.01}^{11}}} \right) \\
   + \left( {\begin{array}{*{20}{c}}
  {12} \\
  {12}
\end{array}} \right)\left( {{3^0}} \right)\left( { - {{0.01}^{12}}} \right) \\
 \]
Here we can consider first five steps only as with the increase in the power of $ - 0.01$, the values become negligible.
\[
   \Rightarrow {\left( {3 - 0.01} \right)^{12}} \\
   = \left( {\begin{array}{*{20}{c}}
  {12} \\
  0
\end{array}} \right)\left( {{3^{12}}} \right)\left( { - {{0.01}^0}} \right) + \left( {\begin{array}{*{20}{c}}
  {12} \\
  1
\end{array}} \right)\left( {{3^{11}}} \right)\left( { - {{0.01}^1}} \right) + \left( {\begin{array}{*{20}{c}}
  {12} \\
  2
\end{array}} \right)\left( {{3^{10}}} \right)\left( { - {{0.01}^2}} \right) + \left( {\begin{array}{*{20}{c}}
  {12} \\
  3
\end{array}} \right)\left( {{3^9}} \right)\left( { - {{0.01}^3}} \right) \\
   + \left( {\begin{array}{*{20}{c}}
  {12} \\
  4
\end{array}} \right)\left( {{3^8}} \right)\left( { - {{0.01}^4}} \right) \\
   = 531441 - 0.12(177147) + 0.0066(59049) - 0.000220(19683) + 0.00000495(6561) \\
   = 531441 - 21257.64 + 389.7234 - 4.330260 + 0.3247695 \\
   \approx 510568.79 \\
 \]
Thus, by using binomial theorem, we can approximate ${\left( {2.99} \right)^{12}}$ as $510568.79$.

Note: We have used the binomial theorem to solve this problem. We know that 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 the Binomial Theorem As we have done in this question.