Summation to Equation

Algebra Level 4

Consider the polynomial representation $h(x)$ of $f(x) = \sum\limits_{n=x}^{x^2} n$ for all positive integers $x$ .

If $h(x)$ has complex roots $a_1, a_2, a_3, a_4$ , then evaluate the sum of the coefficients of the monic quartic polynomial $g(x)$ possessing roots $a_1^2, a_2^2, a_3^2, a_4^2$ .

1 -1 0 None of These 1/2

1 solution

Alex Delhumeau
Mar 9, 2016

Consider some values of $f(x)=\sum\limits_{n=x}^{x^2}n$ .

$f(1)=1$ ; $f(2)=2+3+4=9$ ; $f(3)=3+4+5+6+7+8+9=42$ , etc.

Clearly, $f(x)$ is the sum of the first $x^2$ positive integers minus the sum of the first $x-1$ positive integers. This is equivalent to $(\frac{x^2(x^2+1)}{2}-\frac{(x-1)x}{2})$ , which simplifies to $\frac{x^4+x}{2}$ .

By inspection, two roots of this expression are $-1$ & $0$ . Dividing these out and applying quadratic formula to the resultant polynomial, we find the other two roots as $\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$ .

Squaring all of these, we get $-\frac{1}{2}\pm i\frac{\sqrt{3}}{2},0,$ & $1$ .

Thus we find that $g(x)=ax(x-1)(x+\frac{1}{2}-\frac{i\sqrt{3}}{2})(x+\frac{1}{2}+\frac{i\sqrt{3}}{2})=\frac{ax^4-ax}{2}$ .

The sum of the coefficients, then, no matter what the leading coefficient $a$ may be, $=\boxed{0}$ .

Summation to Equation

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