The least value of n in order that the sum of first n terms of the infinite series `1+3/4+(3/4)^2+(3/4)^3+...`, should differ from the sum of the series by less than `10^-6`, is `(given log2=0.30103,log3=0.47712)`

Options

A. 14
B. 27
C. 53
D. 57

Answer

To find the least value of \( n \) such that the sum of the first \( n \) terms of the infinite series \( 1 + \frac{3}{4} + \left(\frac{3}{4}\right)^2 + \left(\frac{3}{4}\right)^3 + \ldots \) differs from the sum of the series by less than \( 10^{-6} \), we can follow these steps:

### Step 1: Identify the first term and common ratio
The first term \( a \) of the series is \( 1 \) and the common ratio \( r \) is \( \frac{3}{4} \).

### Step 2: Calculate the sum of the infinite series
The sum \( S \) of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
Substituting the values:
\[ S = \frac{1}{1 - \frac{3}{4}} = \frac{1}{\frac{1}{4}} = 4 \]

### Step 3: Write the formula for the sum of the first \( n \) terms
The sum \( S_n \) of the first \( n \) terms of a geometric series is given by:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
Substituting the values:
\[ S_n = \frac{1(1 - \left(\frac{3}{4}\right)^n)}{1 - \frac{3}{4}} = 4(1 - \left(\frac{3}{4}\right)^n) \]

### Step 4: Set up the inequality for the difference
We need the difference between the sum of the infinite series and the sum of the first \( n \) terms to be less than \( 10^{-6} \):
\[ |S - S_n| < 10^{-6} \]
Substituting the values:
\[ |4 - 4(1 - \left(\frac{3}{4}\right)^n)| < 10^{-6} \]
This simplifies to:
\[ |4\left(\frac{3}{4}\right)^n| < 10^{-6} \]
Dividing both sides by 4:
\[ \left(\frac{3}{4}\right)^n < \frac{10^{-6}}{4} = 2.5 \times 10^{-7} \]

### Step 5: Take logarithm on both sides
Taking logarithm (base 10) on both sides:
\[ \log\left(\left(\frac{3}{4}\right)^n\right) < \log(2.5 \times 10^{-7}) \]
Using the property of logarithms:
\[ n \log\left(\frac{3}{4}\right) < \log(2.5) + \log(10^{-7}) \]
Calculating \( \log(10^{-7}) = -7 \) and \( \log(2.5) \):
\[ \log(2.5) = \log\left(\frac{5}{2}\right) = \log(5) - \log(2) \approx 0.6990 - 0.3010 = 0.3980 \]
Thus:
\[ \log(2.5 \times 10^{-7}) = 0.3980 - 7 = -6.602 \]
So we have:
\[ n \log\left(\frac{3}{4}\right) < -6.602 \]

### Step 6: Calculate \( \log\left(\frac{3}{4}\right) \)
Using the property of logarithms:
\[ \log\left(\frac{3}{4}\right) = \log(3) - \log(4) = 0.47712 - 0.60206 = -0.12494 \]
Thus:
\[ n (-0.12494) < -6.602 \]
Dividing both sides by \(-0.12494\) (note that this reverses the inequality):
\[ n > \frac{-6.602}{-0.12494} \approx 52.8 \]

### Step 7: Find the least integer value of \( n \)
Since \( n \) must be an integer, we round up:
\[ n \geq 53 \]

### Final Answer
The least value of \( n \) is \( 53 \).