Find the sum to first n terms :
`1+2/3 + 3/(3^2) + 4/(3^3)`+….

Options

A. a)`3/4 - (3+2n)/(4(3^(n-1)))`
B. b)`9/4 - 3/4 [(3+2n)/(3^n)]`
C. c)`2n - (1/n)^2`
D. d)none of these

Answer

To find the sum of the first n terms of the series \( S_n = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots \), we can follow these steps:

### Step 1: Identify the General Term
The general term of the series can be expressed as:
\[ T_n = \frac{n}{3^{n-1}} \]
This means that the first term \( T_1 = 1 \), the second term \( T_2 = \frac{2}{3} \), the third term \( T_3 = \frac{3}{3^2} \), and so on.

### Step 2: Write the Sum of the First n Terms
The sum of the first n terms can be written as:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k}{3^{k-1}} \]

### Step 3: Use the Formula for the Sum of a Series
To find the sum \( S_n \), we can use the formula for the sum of a series involving \( k \) and a geometric progression. The formula for the sum of the first n terms of the series \( \sum_{k=1}^{n} kx^{k-1} \) is given by:
\[ \sum_{k=1}^{n} kx^{k-1} = \frac{1 - (n+1)x^n + nx^{n+1}}{(1-x)^2} \]
In our case, \( x = \frac{1}{3} \).

### Step 4: Substitute x into the Formula
Substituting \( x = \frac{1}{3} \) into the formula gives:
\[ S_n = \frac{1 - (n+1)\left(\frac{1}{3}\right)^n + n\left(\frac{1}{3}\right)^{n+1}}{\left(1 - \frac{1}{3}\right)^2} \]
Calculating \( 1 - \frac{1}{3} = \frac{2}{3} \), we have:
\[ S_n = \frac{1 - (n+1)\left(\frac{1}{3}\right)^n + n\left(\frac{1}{3}\right)^{n+1}}{\left(\frac{2}{3}\right)^2} \]

### Step 5: Simplify the Expression
Now, simplifying the denominator:
\[ \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]
Thus,
\[ S_n = \frac{9}{4} \left( 1 - (n+1)\left(\frac{1}{3}\right)^n + n\left(\frac{1}{3}\right)^{n+1} \right) \]

### Step 6: Final Expression
The final expression for the sum of the first n terms is:
\[ S_n = \frac{9}{4} \left( 1 - \frac{(n+1)}{3^n} + \frac{n}{3^{n+1}} \right) \]