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 $P_n(x)=1+2x+3x^2+....+(n+1)x^n$
$\therefore$ $x{P}_n\left(x\right)=x+2x^2+.....+n{x}^n+\left(n+1\right)x^{n+1}$
$\therefore \left(1-x\right)P_n\left(x\right)=1+x+x^2+....+x^n-\left(n+1\right)x^{n+1}$
$=\frac{1\left[1-x^{n+1}\right]}{1-x}-\left(n+1\right)x^{n+1}$
= $\frac{1-x^{n+1}-\left(n+1\right)x^{n+1}+\left(n+1\right)x^{n+2}}{1-x}$
= $\frac{1-\left(n+2\right)x^{n+1}+\left(n+1\right)x^{n+2}}{1-x}$
$\therefore P_n\left(x\right)$ = $\frac{1-\left(n-2\right)x^{n+1}+\left(n+1\right)x^{n+2}}{{\left(1-x\right)}^2}$
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 $f(x)$.