Question

Let $n$ be the an odd integer. Then the number of real roots of the polynomial equation . $P_n(x) = 1 + 2x + 3x^2 + ..... + (n + 1)x^n$ is

• A. $0$
• B. $n$
• C. $1$
• D. none of these

Answer 1

Answer: (C)

When $x > 0, P_n(x) > 0$ and
$\therefore \, P_n(x) = 0$ can have no +ve real root.
Again $P_n(x) = 1 + 2x + 3x^2 + .... + (n +1) \, x^n$
$\therefore $ $x\,P_{n}\left(x\right)= x +2\,x^{2} + ..... + n\, x^{n } + \left(n +1\right) x^{n+1} $
$\therefore \, \left(1-x\right) P_{n} \left(x\right) =1+x+x^{2} + .... + x^{n } - \left(n+1\right) x^{n+1}$
$= \frac{1\left[1-x^{n+1}\right]}{1-x} - \left(n+1\right)x^{n+1} $
= $\frac{1-x^{n+1} - \left(n+1\right)x^{n+1} +\left(n + 1\right)x^{n+2}}{1-x} $
= $\frac{1-\left(n +2\right)x^{n+1}+\left(n+1\right)x^{n+2}}{1-x}$
$\therefore \, P_{n}\left(x\right) $ = $\frac{1-\left(n-2\right)x^{n+1} + \left(n+1\right)x^{n+2}}{\left(1-x\right)^{2}}$
For -ve values of $x, P_n(x)$ will vanish whenever
$ f(x) = 1 - (n + 2)x^{n +1} + (n + 1) \,x^{n+ 2} = 0$
Further $f(-x) = 1 - (n + 2) (- 1)^{n+1} \, x^{n+1}$
$ + (n + 1)(- 1)^{n + 2} \, x^{n+2}$
If $n$ is odd, then there is one change of sign in this expression and
$\therefore $ there is one negative real root of $f(x)$.