Chapter 12.6, Problem 37E

(a)

To determine

To find:the expression the $1.01$ as a binomial.

Answer to Problem 37E

The binomial expansion is $\left(1+0.01\right)$ .

Explanation of Solution

Given:

  ${\left(1.01\right)}^4$ .

Concept used:

Binomial theorem:

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\frac{n\left(n-1\right)}{1.2}{x}^{n-2}{y}^2+\frac{n\left(n-1\right)\left(n-2\right)}{\mathrm{1.2.4}}{x}^{n-3}{y}^3+\mathrm{....}{y}^n$ .

Calculation:

Consider the value:

  $1.01$ .

The binomial expansion of the value $1.01$ is:

  $\left(1+0.01\right)$ .

Hence, the binomial expansion is $\left(1+0.01\right)$ .

(b)

To determine

To find:the binomial representation $1.01$ found in the part $a$ and the binomial theorem to calculate the value of ${\left(1.01\right)}^4$ to eight decimals places.

Answer to Problem 37E

The value of ${\left(1.01\right)}^4$ correct to eight decimal digits.

Explanation of Solution

Given: ${\left(1.01\right)}^4$ .

Concept used:

Binomial theorem:

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\frac{n\left(n-1\right)}{1.2}{x}^{n-2}{y}^2+\frac{n\left(n-1\right)\left(n-2\right)}{\mathrm{1.2.4}}{x}^{n-3}{y}^3+\mathrm{....}{y}^n$ .

Calculation:

Consider the equation:

  ${\left(1.01\right)}^4={\left(1+0.01\right)}^4$ .

The binomial theorem is as follows:

If $n$ is positive integer, then the following is true.

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\frac{n\left(n-1\right)}{1.2}{x}^{n-2}{y}^2+\frac{n\left(n-1\right)\left(n-2\right)}{\mathrm{1.2.4}}{x}^{n-3}{y}^3+\mathrm{....}{y}^n$ .

The expansion of ${\left(1+0.01\right)}^4$ will have five terms.

Find the first three terms using the sequence as:

  $1,\frac{4}{1}\text{ or }4\text{ and }\frac{4.3}{1.2}\text{ or 6}\text{.}$

Then use symmetry to find the following terms:

  $\begin{array}{l}{\left(1.01\right)}^4={\left(1+0.01\right)}^4.\\ \Rightarrow 1^4+4{\left(1\right)}^3\left(0.01\right)+\frac{4.3}{1.2}{\left(1\right)}^2{\left(0.01\right)}^2+\frac{\mathrm{4.3.2}}{\mathrm{1.2.3}}\left(1\right){\left(0.01\right)}^3+{\left(0.01\right)}^4.\\ \Rightarrow 1+0.04+0.0006+0.000004+\mathrm{0.00000001.}\\ \Rightarrow \mathrm{1.04060401.}\end{array}$

Hence, the above is the value of ${\left(1.01\right)}^4$ correct to eight decimal digits.

(c)

To determine

To find:the estimation of ${\left(1.01\right)}^4$ to eight decimal using calculator and compare the value to the following in part $b$ .

Answer to Problem 37E

  $1.04060401$ is same as that on part $\left(b\right)$ .

Explanation of Solution

Given:

  ${\left(1.01\right)}^4$ .

Concept used:

Binomial theorem:

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\frac{n\left(n-1\right)}{1.2}{x}^{n-2}{y}^2+\frac{n\left(n-1\right)\left(n-2\right)}{\mathrm{1.2.4}}{x}^{n-3}{y}^3+\mathrm{....}{y}^n$

Calculation:

The value of ${\left(1.01\right)}^4$ using calculator corrected to eight decimal digits is:

  $\mathrm{1.04060401.}$

On comparing the values with that on part $\left(b\right)$ both come up to be equal.

Hence, $1.04060401$ is same as that on part $\left(b\right)$ .