Let A be `nxxn` matrix given by
`A=[(a_(11),a_(12),a_(13)……a_(1n)),(a_(21),a_(22),a_(23)…a_(2n)),(vdots, vdots, vdots),(a_(n1),a_(n2),a_(n3).a_("nn"))]`
Such that each horizontal row is arithmetic progression and each vertical column is a geometrical progression. It is known that each column in geometric progression have the same common ratio. Given that `a_(24)=1,a_(42)=1/8` and `a_(43)=3/16`
Let `d_(i)` be the common difference of the elements in with row then `sum_(i=1)^(n)d_(i)` is

Options

A. `n`
B. `1/2-1/(2^(n+1))`
C. `1-1/(2^(n))`
D. `(n+1)/(2^(n))`

Answer

To solve the problem step by step, we will analyze the given matrix \( A \) and use the information provided to find the required sum of the common differences of the rows.

### Step 1: Understanding the Matrix Structure
The matrix \( A \) is an \( n \times n \) matrix where:
- Each row is an arithmetic progression (AP).
- Each column is a geometric progression (GP) with the same common ratio \( r \).

### Step 2: Given Values
We are given:
- \( a_{24} = 1 \)
- \( a_{42} = \frac{1}{8} \)
- \( a_{43} = \frac{3}{16} \)

### Step 3: Finding \( a_{44} \)
Since \( a_{42} \) and \( a_{43} \) are in the same row (4th row), we can find the common difference \( d_4 \) of the 4th row:
\[ d_4 = a_{43} - a_{42} = \frac{3}{16} - \frac{1}{8} = \frac{3}{16} - \frac{2}{16} = \frac{1}{16} \]
Now, we can find \( a_{44} \):
\[ a_{44} = a_{43} + d_4 = \frac{3}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4} \]

### Step 4: Finding the Common Ratio \( r \)
Using the values of \( a_{24} \) and \( a_{44} \), we can find the common ratio \( r \):
\[ a_{24} \cdot r^2 = a_{44} \]
Substituting the known values:
\[ 1 \cdot r^2 = \frac{1}{4} \implies r^2 = \frac{1}{4} \implies r = \frac{1}{2} \]

### Step 5: Finding Other Elements in the Matrix
Now we can find other elements in the 2nd column:
- \( a_{32} = a_{42} \cdot \frac{1}{r} = \frac{1}{8} \cdot 2 = \frac{1}{4} \)
- \( a_{22} = a_{32} \cdot \frac{1}{r} = \frac{1}{4} \cdot 2 = \frac{1}{2} \)
- \( a_{12} = a_{22} \cdot \frac{1}{r} = \frac{1}{2} \cdot 2 = 1 \)

Now for the 3rd column:
- \( a_{43} = \frac{3}{16} \)
- \( a_{33} = a_{43} \cdot \frac{1}{r} = \frac{3}{16} \cdot 2 = \frac{3}{8} \)
- \( a_{23} = a_{33} \cdot \frac{1}{r} = \frac{3}{8} \cdot 2 = \frac{3}{4} \)
- \( a_{13} = a_{23} \cdot \frac{1}{r} = \frac{3}{4} \cdot 2 = \frac{3}{2} \)

### Step 6: Finding Common Differences \( d_i \)
Now we can calculate the common differences for each row:
- \( d_1 = a_{13} - a_{12} = \frac{3}{2} - 1 = \frac{1}{2} \)
- \( d_2 = a_{23} - a_{22} = \frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4} \)
- \( d_3 = a_{33} - a_{32} = \frac{3}{8} - \frac{1}{4} = \frac{3}{8} - \frac{2}{8} = \frac{1}{8} \)
- \( d_4 = \frac{1}{16} \) (calculated earlier)

### Step 7: Summing the Common Differences
Now we can sum the common differences:
\[ \sum_{i=1}^{n} d_i = d_1 + d_2 + d_3 + d_4 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} \]

### Step 8: Calculating the Sum
To calculate the sum:
\[ \sum_{i=1}^{n} d_i = \frac{8}{16} + \frac{4}{16} + \frac{2}{16} + \frac{1}{16} = \frac{15}{16} \]

### Final Answer
Thus, the sum of the common differences is:
\[ \sum_{i=1}^{n} d_i = 1 - \frac{1}{2^n} \]