If total number of runs secored is n matches is `(n+1)/(4) (2^(n+1) -n-2) ` where ` n ge 1` , and the runs scored in the
` k^(th)` match is given by `k.2^(n+1 - k) `, where ` 1 le k le n, n ` is

Options

A. (a)8
B. (b)9
C. (c)7
D. (d)6

Answer

To solve the problem step by step, we need to analyze the given expressions for the total runs scored in \( n \) matches and the runs scored in the \( k^{th} \) match.

### Step 1: Understand the given expressions
The total number of runs scored in \( n \) matches is given by:
\[ S_n = \frac{n+1}{4} (2^{n+1} - n - 2) \]
The runs scored in the \( k^{th} \) match is given by:
\[ R_k = k \cdot 2^{n+1 - k} \]
where \( 1 \leq k \leq n \).

### Step 2: Set up the summation for total runs
We can express the total runs scored \( S_n \) as a summation of the runs scored in each match:
\[ S_n = \sum_{k=1}^{n} R_k = \sum_{k=1}^{n} k \cdot 2^{n+1 - k} \]

### Step 3: Simplify the summation
We can factor out \( 2^{n+1} \) from the summation:
\[ S_n = 2^{n+1} \sum_{k=1}^{n} k \cdot 2^{-k} \]
Now, we need to evaluate the summation \( \sum_{k=1}^{n} k \cdot 2^{-k} \).

### Step 4: Use the formula for the summation
The summation \( \sum_{k=1}^{n} k x^k \) can be evaluated using the formula:
\[ \sum_{k=1}^{n} k x^k = x \frac{d}{dx} \left( \sum_{k=0}^{n} x^k \right) \]
The geometric series \( \sum_{k=0}^{n} x^k = \frac{1-x^{n+1}}{1-x} \).

### Step 5: Differentiate the geometric series
Differentiating gives:
\[ \frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) = \frac{(1-x^{n+1})'}{1-x} + (1-x^{n+1}) \frac{(1-x)'}{(1-x)^2} \]
Calculating the derivatives and simplifying leads to:
\[ \sum_{k=1}^{n} k x^k = \frac{x(1 - (n+1)x^n + n x^{n+1})}{(1-x)^2} \]
Substituting \( x = \frac{1}{2} \):
\[ \sum_{k=1}^{n} k \cdot 2^{-k} = \frac{\frac{1}{2} \left( 1 - (n+1)\left(\frac{1}{2}\right)^n + n\left(\frac{1}{2}\right)^{n+1} \right)}{\left(1 - \frac{1}{2}\right)^2} \]

### Step 6: Substitute back into \( S_n \)
Now substitute this back into the expression for \( S_n \):
\[ S_n = 2^{n+1} \cdot \text{(result from previous step)} \]

### Step 7: Equate to the given expression
Now we equate this expression to the given total runs expression:
\[ \frac{n+1}{4} (2^{n+1} - n - 2) \]
This leads us to solve for \( n \).

### Step 8: Solve for \( n \)
After simplifying and rearranging, we find that:
\[ n + 1 = 8 \implies n = 7 \]

### Conclusion
Thus, the required answer is:
\[ \boxed{7} \]