JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Sequences and Series

Observe the pattern: each term can be written as $1-{\left(\frac{2}{3}\right)}^n$ for $n=1,2,3,\dots$

Verify:

• $1-\frac{2}{3}=\frac{1}{3}$ ✓
• $1-{\left(\frac{2}{3}\right)}^2=1-\frac{4}{9}=\frac{5}{9}$ ✓
• $1-{\left(\frac{2}{3}\right)}^3=1-\frac{8}{27}=\frac{19}{27}$ ✓

Sum of first 100 terms:
$S={\sum }_{n=1}^{100}\left[1-{\left(\frac{2}{3}\right)}^n\right]$
$=100-{\sum }_{n=1}^{100}{\left(\frac{2}{3}\right)}^n$
$=100-\frac{\frac{2}{3}\left(1-{\left(\frac{2}{3}\right)}^{100}\right)}{1-\frac{2}{3}}$
$=100-2\left(1-{\left(\frac{2}{3}\right)}^{100}\right)$
$=98+2{\left(\frac{2}{3}\right)}^{100}$

Since ${\left(\frac{2}{3}\right)}^{100}$ is a very small positive number, $S$ is slightly greater than 98.

The greatest integer less than or equal to $S$ is 98.