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

$$ p_n(x)=1+2x+3x^2+....+(n+1)x^n\\ x.p_n(x)=x+2x^2+....nx^n+(n+1)x^{n+1}\\ p(x)[1-x]=1+x+x^2+....+x^n-(n+1)x^{n+1}\\ p(x)=\frac{1+x+x^2+....+x^{n}}{1-x}-\frac{(n+1)x^{n+1}}{1-x}=\frac{x^{n+1}-1}{x-1}\frac{1}{1-x}-\frac{(n+1)x^{n+1}}{1-x}\\ =\frac{x^{n+1}-1-(n+1)(x-1)x^{n+1}}{-(x-1)^2}=\frac{x^{n+1}-1+(n+1)x^{n+1}-(n+1)x^{n+2}}{-(x-1)^2}\\ =\frac{(n+2)x^{n+1}-(n+1)x^{n+2}-1}{-(x-1)^2}=\frac{1-(n+2)x^{n+1}+(n+1)x^{n+2}}{(x-1)^2}=0\\ \implies \boxed{(n+2)x^{n+1}-(n+1)x^{n+2}=1} $$

I think I am stuck with my attempt, how can I find the real solutions ?

  • $\begingroup$ Uhhh ... did you forget to post something? This post seems half-finished. What is your question, for that matter? $\endgroup$ – Eevee Trainer Apr 15 at 6:55
  • $\begingroup$ @Eevee Trainer pls check i hv edited OP $\endgroup$ – ss1729 Apr 15 at 7:37
  • $\begingroup$ The last expression is not correct; at least, the $1$ is no more here $\endgroup$ – Claude Leibovici Apr 15 at 8:00
  • $\begingroup$ @ClaudeLeibovici thanx . pls check hope i have corrected it. $\endgroup$ – ss1729 Apr 15 at 8:33
  • $\begingroup$ This is correct now. Thanks for fixing it. Now, follow Gerry Myerson's hit.Cheers :-) $\endgroup$ – Claude Leibovici Apr 15 at 9:04

$$ p(x)=1+2x+3x^2+\ldots+(n+1)x^n;\\ xp(x)=x+2x^2+3x^3+\ldots+(n+1)x^{n+1};\\ (1-x)p(x)=1+x+x^2+\ldots+x^n-(n+1)x^{n+1}=\frac{1-x^{n+1}}{1-x}-(n+1)x^{n+1}\\ =\frac{1-x^{n+1}-(1-x)(n+1)x^{n+1}}{1-x}=0.\\ \Rightarrow1-x^{n+1}-(1-x)(n+1)x^{n+1}=0 $$ Note that $x=1$ is clearly not a solution of the original equation. As a result, the above equation is actually equivalant to your original equation, apart from having an extra root $x=1$.

Here is the equation we are trying to solve: $$ 1-x^{n+1}-(1-x)(n+1)x^{n+1}\equiv (n+1)x^{n+2}-(n+2)x^{n+1}+1=0\\ \Leftrightarrow(n+1)x^{n+2}=(n+2)x^{n+1}-1 $$ Since the original equation has no roots for $x\geq 0$, we now focus on the case $x<0$. When $x<0$, LHS is decreasing toward $-\infty$; RHS is increasing towards $\infty$. LHS>RHS for $x=0$. This tells us that the original equation has only one root.

  • $\begingroup$ I understand ur first statement, no solutions for $x\geq0$ as $p(x)>0$. But could u pls explain the later part more. $\endgroup$ – ss1729 Apr 15 at 15:13
  • $\begingroup$ so u mean .. as LHS is strictly decreasing towards $-\infty$ and RHS is strictly increasing towards $+\infty$, both curves should intersect at some point, thus have only one root, right ?. But, how can we find the value of the real root ? $\endgroup$ – ss1729 Apr 15 at 18:00
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    $\begingroup$ You can find it using WolframAlpha. It's value should tend to $1$ as $x \to \infty$. However, you know, algebraic equation of degree $5$ or higher doesn't have a formula. $\endgroup$ – Holding Arthur Apr 15 at 22:02

You have found that $p(x)$ is a rational function. It is zero if and only if its numerator is zero. Its numerator is a polynomial of the form $Ax^{n+2}+Bx^{n+1}+C$, where $A,B,C$ are constants. The derivative of that numerator is of the form $Dx^{n+1}+Ex^n=x^n(Dx+E)$ for some constants $D,E$. It's easy to see how many real roots that derivative has. Then Rolle's Theorem tells you something about how many real roots the original numerator has. Can you fill in the details and draw a conclusion?

  • $\begingroup$ could u please add more details as I am having trouble following ur hint. $\endgroup$ – ss1729 Apr 15 at 17:40
  • $\begingroup$ You'll have to do better than that. Where are you having trouble? Let's start at the beginning. Do you know what a rational function is? $\endgroup$ – Gerry Myerson Apr 15 at 23:27
  • $\begingroup$ I guess not. OK, a rational function is a quotient of two polynomials. Any other part of my answer giving you trouble? I can't help you, if you won't tell me what's wrong. $\endgroup$ – Gerry Myerson Apr 17 at 10:27
  • $\begingroup$ extremely sorry for the delay. I understand what a rational function is. and here it is of the form $Ax^{n+2}+Bx^{n+1}+C$ which can be seen from my final equation. And its derivative $Dx^{n+1}+Ex^n=x^n(Dx+E)$.So far its clear. And rolles theorem tells u that if the function has two roots then there must be atleast one stationary point where the derivative is zero in between them. But, how do I use it here to reach the desired conclusion ? $\endgroup$ – ss1729 Apr 17 at 13:18
  • $\begingroup$ $A=n+1,B=-n-2,C=1,D=(n+1)(n+2),E=-(n+1)(n+2)\implies E=-D$. $p(x)=Ax^{n+2}+Bx^{n+1}+C$ and $p'(x)=Dx^{n+1}+Ex^n=x^n(Dx+E)$.Now, $p'(x)=0\implies x=0$ or $x=-E/D=-1$, right ?.Now how do I use Rolle's theorem ? $\endgroup$ – ss1729 Apr 17 at 13:27

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