Question

Find the value of $\frac{1}{47}$ to four places of decimals.

Answer 1

$\frac{1}{47}=(47{)}^{-1}=(7^2-2{)}^{-1}=\frac{1}{49}{(1-\frac{2}{7^2})}^{-1}$

$=\frac{1}{49}(1+\frac{2}{7^2}+\frac{2^3}{7^4}+3.\frac{2^4}{7^6}+.....)$

$=\frac{1}{7^2}+\frac{2}{7^4}+\frac{2^3}{7^6}+......$

To obtain the values of several terms we proceed as follow

$\frac{1}{7^2}=0.02040816$

$\frac{1}{7^4}=0.0004164$

$\frac{1}{7^6}=0.00000849$

Since, $\frac{1}{7^6}$ starting with five ciphers, so it will not contribute to result.

$\therefore \frac{1}{47}=0.02040816+2\times 0.0004164$

$=0.02040816+0.0008328$

$=0.02124096$

So, $\frac{1}{47}=0.0212$ upto four decimal places.