Question

The least positive integer $n$ such that $1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - ....... - \dfrac{2}{{{3^{n - 1}}}} < \dfrac{1}{{100}}$
A.$4$
B.$5$
C.$6$
D.$7$

Answer 1

Here we substitute value from each option and check if the value on LHS is less than the value on RHS of the inequality. We can solve this question using Trial and error method.

Complete step-by-step answer:
We are given $1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - ....... - \dfrac{2}{{{3^{n - 1}}}} < \dfrac{1}{{100}}$
We first find the value of RHS to which we will compare the sum of terms on LHS.
We have RHS as $\dfrac{1}{{100}}$which can be written in the decimal form as $0.01$
So, we take the value in RHS as $0.01$.
Now we carefully assess each option.
Option A.
Here the value of $n = 4$.
We substitute the value of $n = 4$ in the LHS of the equation.

$$\Rightarrow 1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - ....... - \dfrac{2}{{{3^{4 - 1}}}} = 1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - \dfrac{2}{{{3^3}}} \\\ \Rightarrow 1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - ....... - \dfrac{2}{{{3^3}}} = 1 - \dfrac{2}{3} - \dfrac{2}{9} - \dfrac{2}{{27}} \\\$$

Now we take LCM

$$\Rightarrow \dfrac{{27 - 2 \times 9 - 2 \times 3 - 2}}{{27}} \\\ \Rightarrow \dfrac{{27 - 18 - 6 - 2}}{{27}} \\\ \Rightarrow \dfrac{1}{{27}} \\\$$

$\Rightarrow 1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - \dfrac{2}{{{3^3}}} = \dfrac{1}{{27}}$ … (1)
Calculating the value of $\dfrac{1}{{27}}$ we get $0.03$
Now since $0.03 > 0.01$
Therefore, option A is rejected.
Option B.
Here the value of $n = 5$.
We substitute the value of $n = 5$ in the LHS of the equation.
$\Rightarrow 1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - ....... - \dfrac{2}{{{3^{5 - 1}}}} = 1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - \dfrac{2}{{{3^3}}} - \dfrac{2}{{{3^4}}}$
We know from equation (1) that $1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - \dfrac{2}{{{3^3}}} = \dfrac{1}{{27}}$ , substitute the value in above equation
$\Rightarrow 1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - \dfrac{2}{{{3^3}}} - \dfrac{2}{{{3^4}}} = \dfrac{1}{{27}} - \dfrac{2}{{81}}$
Now we take LCM

$$\Rightarrow \dfrac{{1 \times 3 - 2}}{{81}} \\\ \Rightarrow \dfrac{{3 - 2}}{{81}} \\\ \Rightarrow \dfrac{1}{{81}} \\\$$

$\Rightarrow 1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - \dfrac{2}{{{3^3}}} - \dfrac{2}{{{3^4}}} = \dfrac{1}{{81}}$ … (2)
Calculating the value of $\dfrac{1}{{81}}$ we get $0.012$
Now since $0.012 > 0.010$
Therefore, option B is rejected.
Option C.
Here the value of $n = 6$.
We substitute the value of $n = 6$ in the LHS of the equation.
$\Rightarrow 1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - ....... - \dfrac{2}{{{3^{6 - 1}}}} = 1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - \dfrac{2}{{{3^3}}} - \dfrac{2}{{{3^4}}} - \dfrac{2}{{{3^5}}}$
We know from equation (2) that $1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - \dfrac{2}{{{3^3}}} - \dfrac{2}{{{3^4}}} = \dfrac{1}{{81}}$ , substitute the value in above equation
$\Rightarrow 1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - \dfrac{2}{{{3^3}}} - \dfrac{2}{{{3^4}}} - \dfrac{2}{{{3^5}}} = \dfrac{1}{{81}} - \dfrac{2}{{273}}$
Now we take LCM

$$\Rightarrow \dfrac{{1 \times 3 - 2}}{{273}} \\\ \Rightarrow \dfrac{{3 - 2}}{{273}} \\\ \Rightarrow \dfrac{1}{{273}} \\\$$

$\Rightarrow 1 - \dfrac{2}{3} - \dfrac{2}{{{3^2}}} - \dfrac{2}{{{3^3}}} - \dfrac{2}{{{3^4}}} - \dfrac{2}{{{3^5}}} = \dfrac{1}{{273}}$ … (3)
Calculating the value of $\dfrac{1}{{273}}$ we get $0.003$
Now since $0.003 < 0.01$
Therefore, option C is accepted.
Therefore, least positive integer $n = 6$
Now since option D has n greater than 6, we will not check for option D as we have to find the least value of n and we have the least value as 6.
So, option C is correct.

Note: Students many times make the mistake of calculating the LCM of all the values again in each step which makes our calculation complex, we should always use the previous deductions to solve further parts. Also, while comparing the decimal values, always keep in mind the position of decimal.