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For all real number $x$ which don't make denominators 0, a following equation always holds. $$\frac{4}{x^2 - 1} + \frac{8}{x^2 - 4} +\frac{12}{x^2 - 9} +\dots+ \frac{40}{x^2 - 100}\\ = k \left(\frac{1}{(x-1)(x+10)} +\frac{1}{(x-2)(x+9)} +\dots + \frac{1}{(x-10)(x+1)} \right)$$ What is the constant $k$?

Calculators can not be used to solve this problem.

JJH
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    To get answers that help you learn how to solve problems like this, you need to tell us a few things about the tools available to you. Are you familiar with any of the following: limits, residues, partial fractions,...? Your last sentence in particular hints at this being copy/pasted from either your textbook or perhaps some test, or a contest, where participants are constrained in this way. This adds some urgency to the request: What's the source of this problem? – Jyrki Lahtonen Mar 19 '18 at 08:18
  • I want algerbra level answer like partial fractions. How can this problem be solved using residues? I can't imagine. The last sentence was written to prevent from using sigma formulas of sequence and calculators that have functions of sigma symbolic calculation. The Source of this problem is math workbook of high school in south korea. I think that the problem is in the identity chapter of the subject math 1. – JJH Mar 19 '18 at 10:59
  • if x=0, the identity can be calculated by calculators, symbolic calculators and sigma formulas. Those are not solution. – JJH Mar 19 '18 at 11:17

3 Answers3

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If you already know that the identity holds, just multiply both sides by $x^2$ and take the limit as $x\to +\infty$. Then we obtain $$(4+8+12+\dots+ 40)=10k\implies k=\frac{4(1+2+3+\dots+10)}{10}=\frac{4\frac{10\cdot 11}{2}}{10}=22.$$ P.S. More generally, now that you got the idea, show that for any positive integer $n$ (and for any real $x\not\in \{\pm j: j=1,2,\dots,n\}$), $$4\sum_{j=1}^n \frac{j}{x^2-j^2}=k_n\sum_{j=1}^n \frac{1}{(x-j)(x+n+1-j)}\tag{1}$$ implies that $k_n=2(n+1)$. In order to show that the identity (1) holds, note that $$4\sum_{j=1}^n \frac{j}{x^2-j^2}=2\sum_{j=1}^n \frac{1}{x-j}-2\sum_{j=1}^n \frac{1}{x+j}$$ and $$2(n+1)\sum_{j=1}^n \frac{1}{(x-j)(x+n+1-j)}= 2\sum_{j=1}^n \frac{1}{x-j}-2\sum_{j=1}^n \frac{1}{x+(n+1-j)}.$$

Robert Z
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Let $x=0$, $$ LHS = \sum_{i=1}^{11} \frac{4*i}{-i^2} = -4 \sum_{i=1}^{10} \frac{1}{i}.$$

$$ RHS = k \sum_{i=1}^{10} \frac{1}{(-i)(11-i)}.$$

Notice that $$ \frac{1}{(-i)(11-i)} = \frac{1}{11}(-\frac{1}{i} - \frac{1}{11-i}).$$

Therefore \begin{equation*} \begin{aligned} RHS & = \frac{k}{11} \sum_{i=1}^{10} (-\frac{1}{i} - \frac{1}{11-i}) = -\frac{k}{11} (\sum_{i=1}^{10} \frac{1}{i} + \sum_{i=1}^{10} \frac{1}{11-i}) \\ & = -\frac{k}{11} (\sum_{i=1}^{10} \frac{1}{i} + \sum_{j=1}^{10} \frac{1}{j}) = -\frac{2}{11}k \sum_{i=1}^{10} \frac{1}{i} . \end{aligned} \end{equation*}

Finally we have $-4 = -\frac{2}{11}k$ and $k=22$.

sbandjohn
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Clear all denominators by multiplying both sides by: $$ (x - 1)(x + 1)(x - 2)(x + 2) \cdots (x - 10)(x + 10) $$ Each of the 10 terms in the LHS would have a factor of $(x - 1)$, except the first term. Likewise, each of the 10 terms in the RHS would have a factor of $(x - 1)$, except the first. So after evaluating at $x = 1$, all terms would vanish except the first term from both sides of the equation. Thus, we obtain: $$ 4\color{red}{(1 - 2)(1 + 2) \cdots (1 - 10)}(1 + 10) = k(1 + 1)\color{red}{(1 - 2)(1 + 2) \cdots (1 - 10)} $$ Cancelling the common red factors, we obtain: $$ 4 \cdot 11 = k \cdot 2 \iff k = 22 $$

Adriano
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