If `I_(n)=int_(0)^(1)(1+x+x^(2)+....+x^(n-1))(1+3x+5x^(2)+....+(2n-3)x^(n-2)+(2n-1)x^(n-1))dx,n in N,` then the value of `sqrt(I_(9))` is (a) 3 (b) 6 (c) 9 (d) 12

Options

A. 3
B. 6
C. 9
D. 12

Explanation

To solve the problem, we need to evaluate the integral \( I_n \) defined as:

\[ I_n = \int_0^1 (1 + x + x^2 + \ldots + x^{n-1})(1 + 3x + 5x^2 + \ldots + (2n-1)x^{n-1}) \, dx \]

### Step 1: Simplify the first part of the integral

The first part of the integral, \( 1 + x + x^2 + \ldots + x^{n-1} \), is a geometric series. The sum of a geometric series can be expressed as:

\[ 1 + x + x^2 + \ldots + x^{n-1} = \frac{1 - x^n}{1 - x} \]

### Step 2: Simplify the second part of the integral

The second part, \( 1 + 3x + 5x^2 + \ldots + (2n-1)x^{n-1} \), can be recognized as a polynomial. This series can be expressed as:

\[ \sum_{k=0}^{n-1} (2k + 1)x^k = 1 + 3x + 5x^2 + \ldots + (2n-1)x^{n-1} \]

This can be derived as follows:

\[ = \sum_{k=0}^{n-1} 2kx^k + \sum_{k=0}^{n-1} x^k = 2x \frac{d}{dx} \left( \frac{1 - x^n}{1 - x} \right) + \frac{1 - x^n}{1 - x} \]

### Step 3: Combine the two parts

Now, we need to combine both parts into the integral:

\[ I_n = \int_0^1 \left( \frac{1 - x^n}{1 - x} \right) \left( 1 + 3x + 5x^2 + \ldots + (2n-1)x^{n-1} \right) \, dx \]

### Step 4: Evaluate the integral

To evaluate \( I_n \), we can use the properties of definite integrals and the results from the previous simplifications.

After performing the calculations and integrating, we find that:

\[ I_n = n^2 \]

### Step 5: Find \( \sqrt{I_9} \)

Now, substituting \( n = 9 \):

\[ I_9 = 9^2 = 81 \]

Thus, we find:

\[ \sqrt{I_9} = \sqrt{81} = 9 \]

### Final Answer

The value of \( \sqrt{I_9} \) is \( 9 \).