Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the sum of the roots of the equation

$$(x − 1) + (x − 2)^2 + (x − 3)^3 + ... + (x − 10)^{10} = 0 $$?

When i expand this equation, it become in the power of 10 and its get complicated. Now what i am thinking is the sum of roots will be equal to the sum of coefficents of x^9 .So i just need to evaluate coefficent of x^9 in the term $$(x-10)x^{10}$$. Am in right in thinking?

But is there is any other easier way by which i can calculate?

Thanks in advance.

share|cite|improve this question
Yes, you are right. Take a look at:'s_formulas – Beni Bogosel Apr 16 '12 at 11:46
I would think you'd be after the coeff of $x^9$ in $(x-9)^9+(x-10)^{10}$. – Gerry Myerson Apr 16 '12 at 11:48
Ya,I dint mention it as the coefficent of x^9 in (x-9)^9 will be 1 for sure. – vikiiii Apr 16 '12 at 11:50

Coefficient of $x^9$ is $$1+ {10 \choose 1} (-10) = 1 - 100 = -99$$

The sum of the roots is therefore $99$ Do you know why?

I am adding this explanation on request: Although, I would recommend you have to read more on Vieta's formula.

If $\alpha$ and $\beta$ are roots of a quadratic equation, then $(x-\alpha)(x-\beta) = x^2-(\alpha+\beta)x+\alpha \beta$ and the absolute value of the coefficient of $x$ is the sum of of the roots. (In this case $\alpha+\beta$.

Similarly a tenth degree polynomial, say has roots $\alpha_1, \alpha_2, \dots \alpha_{10}$ , then the polynomial

$$(x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_{10}) = x^{10}-(\sum_{i=1}^{10} \alpha_i)x^9 +\dots +\prod_{i=1}^{10} \alpha_i$$

share|cite|improve this answer
Its right but Can you please explain? – vikiiii Apr 16 '12 at 11:54
@vikiiii I added explanation as you requested. – Kirthi Raman Apr 16 '12 at 12:10
Raman But here we dont know the roots of equation. So how to find their sum. – vikiiii Apr 23 '12 at 2:38
@vikiii the sum of the roots is the absolute value of the coefficient of $x^9$ which is the absolute value of $1+ {10 \choose 1} (-10) = 1 - 100 = -99$ and that absolute value is $99$. For instance what is the sum of the roots of $(x-2)^2+(x-4)^2=0$? which is $2x^2-12x+20=2(x^2-6x+10)=0$. The sum of the roots is the absolute value of the coefficient of $x$ and that is 6. – Kirthi Raman Apr 23 '12 at 11:28

Hint $\: $ Shift $\rm\: x = y+10,\:$ so $\rm\:y^{10} + (y+1)^9 +\cdots\: = y^{10} + y^9 + \cdots\:$ has root sum $\:\!-1,\:$ so $\rm\:x_1+\cdots + x_{10} =\: (y_1\!+\!10)+\cdots+(y_{10}\!+\!10) =\: y_1+\cdots+y_{10}+100 = -1 + 100 = 99.$

share|cite|improve this answer
Can you explain a little bit more? – vikiiii Apr 23 '12 at 2:38
@vikiiii What is proving problematic? – Bill Dubuque Apr 23 '12 at 2:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.