If $\alpha ,\beta $ are the roots of equation $a{ x }^{ 2 }+bx+c=0$, then $\log _{ e }{ a } +2\log _{ e }{ x } -\cfrac { 1 }{ x } \left( \alpha +\beta \right) -\cfrac { 1 }{ 2{ x }^{ 2 } } \left( { \alpha }^{ 2 }+{ \beta }^{ 2 } \right) -\cfrac { 1 }{ 3{ x }^{ 2 } } \left( { \alpha }^{ 3 }+{ \beta }^{ 3 } \right) ....$ is equal to

Using the factorization of a quadratic equation based on its roots alpha and beta, we establish that ax^2 + bx + c = a(x - alpha)(x - beta). Dividing both sides by a and taking the natural logarithm of both sides results in ln(a) + ln(x^2 + bx/c) or more accurately ln(ax^2 + bx + c) = ln(a) + ln(x - alpha) + ln(x - beta). We expand ln(x - alpha) using the Taylor series for ln(1 - y) to get ln(x) + ln(1 - alpha/x) = ln(x) - (alpha/x + alpha^2/2x^2 + alpha^3/3x^2 + ...). Repeating this expansion for ln(x - beta) yields ln(x) - (beta/x + beta^2/2x^2 + beta^3/3x^2 + ...). Adding ln(a) to the sum of both expansions perfectly matches the given series, resulting in the overall sum being ln(ax^2 + bx + c).

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