Let `t_(r)` denote the rth term in the binomial expansion of `(a + 1)^(50)` . If `t_(25) + t_(27) = 125/52t_(26)` , then the sum of 52 all the values of a is:

Options

A. `1//2`
B. `3//2`
C. `2`
D. `5//2`

Answer

To solve the problem, we will follow the steps outlined in the video transcript to derive the necessary values of \( a \) from the given equation involving the binomial expansion.

### Step-by-Step Solution:

1. **Identify the Terms in the Binomial Expansion:**
The r-th term in the binomial expansion of \( (a + 1)^{50} \) is given by:
\[ T_r = \binom{n}{r-1} \cdot (1)^{r-1} \cdot a^{n-(r-1)} = \binom{50}{r-1} \cdot a^{51-r} \]
Therefore, we can express:
- \( T_{25} = \binom{50}{24} a^{26} \)
- \( T_{26} = \binom{50}{25} a^{25} \)
- \( T_{27} = \binom{50}{26} a^{24} \)

2. **Set Up the Given Equation:**
According to the problem, we have:
\[ T_{25} + T_{27} = \frac{125}{52} T_{26} \]
Substituting the expressions for \( T_{25}, T_{26}, \) and \( T_{27} \):
\[ \binom{50}{24} a^{26} + \binom{50}{26} a^{24} = \frac{125}{52} \cdot \binom{50}{25} a^{25} \]

3. **Simplify the Equation:**
We can factor out \( a^{24} \) from the left side:
\[ a^{24} \left( \binom{50}{24} a^2 + \binom{50}{26} \right) = \frac{125}{52} \cdot \binom{50}{25} a^{25} \]
Dividing both sides by \( a^{24} \) (assuming \( a \neq 0 \)):
\[ \binom{50}{24} a^2 + \binom{50}{26} = \frac{125}{52} \cdot \binom{50}{25} a \]

4. **Rearranging the Equation:**
Rearranging gives us:
\[ \binom{50}{24} a^2 - \frac{125}{52} \cdot \binom{50}{25} a + \binom{50}{26} = 0 \]

5. **Substituting Binomial Coefficients:**
Using the values of the binomial coefficients:
\[ \binom{50}{24} = \frac{50!}{24! \cdot 26!}, \quad \binom{50}{25} = \frac{50!}{25! \cdot 25!}, \quad \binom{50}{26} = \frac{50!}{26! \cdot 24!} \]
We can simplify:
\[ \binom{50}{26} = \binom{50}{24} \quad \text{(since } \binom{n}{k} = \binom{n}{n-k}\text{)} \]
Thus, we can substitute and simplify further.

6. **Final Quadratic Equation:**
After simplification, we arrive at a quadratic equation in terms of \( a \):
\[ a^2 - \frac{5}{2} a + 1 = 0 \]

7. **Finding the Roots:**
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = \frac{\frac{5}{2} \pm \sqrt{\left(-\frac{5}{2}\right)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
Simplifying gives us the roots.

8. **Sum of the Roots:**
The sum of the roots \( \alpha + \beta \) is given by \( -\frac{b}{a} = \frac{5}{2} \).

### Conclusion:
The sum of all values of \( a \) is:
\[ \text{Sum of all values of } a = \frac{5}{2} \]