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

Options

A. `log(1+px+qx^(2))`
B. `log(1-px+qx^(2))`
C. `log(1+px-qx^(2))`
D. `log(1-px-qx^(2))`

Answer

To solve the problem, we need to evaluate the infinite series given by:

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

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

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

### Step 2: Express \(\alpha^2 + \beta^2\) and \(\alpha^3 + \beta^3\)
Using the identities:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = p^2 - 2q \]
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) = p((p^2 - 2q) - q) = p(p^2 - 3q) \]

### Step 3: Substitute these into the series
Now we can rewrite the series \(S\) as:
\[ S = px - \frac{(p^2 - 2q)x^2}{2} + \frac{(p^3 - 3pq)x^3}{3} - \ldots \]

### Step 4: Recognize the series as a logarithmic series
The series resembles the Taylor series expansion for \(\log(1 - u)\):
\[ \log(1 - u) = -\left(u + \frac{u^2}{2} + \frac{u^3}{3} + \ldots\right) \]
Thus, we can express \(S\) in terms of logarithms.

### Step 5: Formulate the logarithmic expression
We can express \(S\) as:
\[ S = \log(1 + px + qx^2) \]

### Conclusion
Thus, the final result is:
\[ S = \log(1 + px + qx^2) \]