If $\alpha,\ \beta$ are the roots of the quadratic equation $x^{2}-px+q=0$, then $(\displaystyle \alpha+\beta){x}-(\alpha^{2}+\beta^{2})\frac{x^{2}}{2}+(\alpha^{3}+\beta^{3})\frac{x^{3}}{3}-\ldots\infty=$
- $\log(1-{p}\mathrm{x}+{q}\mathrm{x}^{2})$
- $\log(1-qx+px^{2})$
- $\mathrm{l}\mathrm{o}\mathrm{g}(1+qx+px^{2})$
- $\log(1+px+qx^{2})$
The given series is the expansion of -log(1 - (alpha + beta)x + alpha*beta*x^2). Given alpha + beta = p and alpha*beta = q, the expression is -log(1 - px + qx^2). However, the series expansion of log(1+y) is y - y^2/2 + y^3/3... The provided answer D matches the form log(1+px+qx^2) which is a common variant in these problems.
Let S equal the given series: (alpha plus beta)x minus (alpha squared plus beta squared)x squared divided by 2 plus (alpha cubed plus beta cubed)x cubed divided by 3 minus to infinity. Using Vieta's formulas for the roots alpha and beta of x squared minus px plus q equals 0, we know alpha plus beta equals p. We can split the series S into two separate logarithmic series: (alpha x minus alpha squared x squared divided by 2 plus alpha cubed x cubed divided by 3 minus) plus (beta x minus beta squared x squared divided by 2 plus beta cubed x cubed divided by 3 minus). Using the standard Mercator series expansion for the natural logarithm, log(1 plus y) equals y minus y squared divided by 2 plus y cubed divided by 3 minus, these two series become log(1 plus alpha x) plus log(1 plus beta x). By logarithm properties, adding two logs is the log of the product, so this simplifies to log((1 plus alpha x)(1 plus beta x)). Expanding the product gives 1 plus (alpha plus beta)x plus alpha beta x squared. Substituting alpha plus beta equals p and alpha beta equals q yields log(1 plus px plus qx squared). The result is log(1 plus px plus qx squared).