If `alpha,beta` are the roots of the equation `x^(2)-px+q=0`, then the value of `(alpha+beta)x-((alpha^(2)+beta^(2))/(2))x^(2)+((alpha^(3)+beta^(3))/(3))x^(3)-. . . .` is

Options

A. `log(1-px+qx^(2))`
B. `log(1+px-qx^(2))`
C. `log(1+px+qx^(2))`
D. None of these

Answer

To solve the problem, we need to find the value of the expression:

\[ (\alpha + \beta)x - \frac{(\alpha^2 + \beta^2)}{2}x^2 + \frac{(\alpha^3 + \beta^3)}{3}x^3 - \ldots \]

where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(x^2 - px + q = 0\).

### Step 1: Identify the roots' properties
From Vieta's formulas, we know:
- \(\alpha + \beta = p\)
- \(\alpha \beta = q\)

### Step 2: Express \(\alpha^2 + \beta^2\)
We can express \(\alpha^2 + \beta^2\) in terms of \(p\) and \(q\):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = p^2 - 2q \]

### Step 3: Express \(\alpha^3 + \beta^3\)
Similarly, we can express \(\alpha^3 + \beta^3\) using the identity:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) = p((p^2 - 2q) - q) = p(p^2 - 3q) \]

### Step 4: Substitute these expressions into the series
Now, we can substitute these expressions back into the original series:
\[ p x - \frac{(p^2 - 2q)}{2} x^2 + \frac{p(p^2 - 3q)}{3} x^3 - \ldots \]

### Step 5: Recognize the series as a logarithmic expansion
The series can be recognized as the Taylor series expansion for \(\log(1 + u)\):
\[ \log(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \ldots \]
where \(u\) is a function of \(x\).

### Step 6: Identify \(u\)
We can identify \(u\) as:
\[ u = \alpha x + \beta x + \alpha \beta x^2 = (p x + q x^2) \]
Thus, we can rewrite the series as:
\[ \log(1 + (p x + q x^2)) \]

### Step 7: Final result
Therefore, the value of the original expression is:
\[ \log(1 + p x + q x^2) \]

### Conclusion
The final answer is:
\[ \log(1 + p x + q x^2) \]