Question

The value of $\frac{1}{\sqrt[5]{5e}}$ , upto four places of decimal is $p.$ find the approximate value of $10p$.

Answer 1

$\frac{1}{\sqrt[5]{5e}}=e^{-1/5}$.

$=1+(-\frac{1}{5})+\frac{1}{2!}{(-\frac{1}{5})}^2+\frac{1}{3!}{(-\frac{1}{5})}^3+\frac{1}{4!}{(-\frac{1}{5})}^4+...$.

$=1-\frac{1}{5}+\frac{1}{50}-\frac{1}{6}.\frac{1}{125}+\frac{1}{24}.\frac{1}{625}-...$.

$=1-\frac{2}{10}+\frac{2}{100}-\frac{1}{6}.\frac{8}{1000}+\frac{1}{24}.\frac{16}{10000}-...$

$=1-0.2+0.02-\frac{1}{6}\times 0.008+\frac{1}{3}\times 0.0002-...$

$=1-0.2+0.02-\frac{0.008}{6}+\frac{0.0002}{3}$

$=1-0.2+0.02-0.001333+0.000066$

$=1-0.2+0.02-0.0013$

$p=0.8187$ (Neglecting other terms)

For four places of decimal, write the series upto 5 terms. For five places of decimal, write the series upto 6 terms and so on.