What is the constant k of a following identical equation?

Question

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.

Answer 1

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 $$2(n+1)\sum_{j=1}^n \frac{1}{(x-j)(x+n+1-j)}= 2\sum_{j=1}^n \frac{1}{x-j}-2\sum_{j=1}^n \frac{1}{x+(n+1-j)}.$$

Answer 2

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 $k=22$.

Answer 3

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: $$ 4 \cdot 11 = k \cdot 2 \iff k = 22 $$