If `(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+… and +C_(n)x^(n)`
`Sigma_(r=0)^(50)(C_(r)^(2))/((r+1))=(m!)/((n!)^(2)),` then the value of `(m+n)` is equal to (where `C_(r)` represents `.^(n)C_(r)` )

Options

A. 149
B. 152
C. 155
D. 146

Answer

To solve the problem, we need to evaluate the expression given and find the values of \( m \) and \( n \) such that:

\[ \sum_{r=0}^{50} \frac{C_r^2}{r+1} = \frac{m!}{(n!)^2} \]

where \( C_r = \binom{n}{r} \).

### Step 1: Understanding the Binomial Expansion
The binomial expansion of \( (1+x)^n \) is given by:

\[ (1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]

This can be written in summation notation as:

\[ (1+x)^n = \sum_{r=0}^{n} C_r x^r \]

### Step 2: Integrating the Binomial Expansion
To find the sum \( \sum_{r=0}^{50} \frac{C_r^2}{r+1} \), we can use integration. We integrate \( (1+x)^n \):

\[ \int (1+x)^n \, dx = \frac{(1+x)^{n+1}}{n+1} + C \]

### Step 3: Evaluating the Integral
We can evaluate the integral from 0 to \( x \):

\[ \int_0^x (1+t)^n \, dt = \left[ \frac{(1+t)^{n+1}}{n+1} \right]_0^x = \frac{(1+x)^{n+1} - 1}{n+1} \]

### Step 4: Using the Result of Integration
Now, we can express the integral in terms of the binomial coefficients:

\[ \int_0^x (1+t)^n \, dt = \sum_{r=0}^{n} C_r \frac{x^{r+1}}{r+1} \]

### Step 5: Comparing Coefficients
Now, we need to multiply the two expansions:

1. \( (1+x)^{n+1} \)
2. \( (1+x)^{n} \)

When we multiply these two expansions, we can find the coefficient of \( x^{51} \):

\[ (1+x)^{n+1} \cdot (1+x)^{n} = (1+x)^{2n+1} \]

### Step 6: Coefficient of \( x^{51} \)
The coefficient of \( x^{51} \) in \( (1+x)^{2n+1} \) is given by:

\[ \binom{2n+1}{51} \]

### Step 7: Setting Up the Equation
From the earlier steps, we have:

\[ \sum_{r=0}^{50} \frac{C_r^2}{r+1} = \frac{1}{n+1} \cdot \binom{2n+1}{51} \]

### Step 8: Equating to Given Expression
We know from the problem statement that:

\[ \sum_{r=0}^{50} \frac{C_r^2}{r+1} = \frac{m!}{(n!)^2} \]

### Step 9: Finding \( m \) and \( n \)
From the equation:

\[ \frac{1}{n+1} \cdot \binom{2n+1}{51} = \frac{m!}{(n!)^2} \]

We can deduce that \( m = 2n + 1 \) and \( n = 51 \).

### Step 10: Calculating \( m+n \)
Thus, we have:

\[ m = 101 \quad \text{and} \quad n = 51 \]

So,

\[ m+n = 101 + 51 = 152 \]

### Final Answer
The value of \( m+n \) is:

\[ \boxed{152} \]