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Solve: $$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$

This is taken from one of the TAU entry tests (I have one in 2 weeks :) )

So, I don't really recognize anything speical here except that there's some similiraty between some pairs of denominators, my intuation is telling me to just multipy everything and try to solve but it looks like a big dead polynomial.

Is there an elegent way to solve this?

Solution:

$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$

$$\frac{1}{x+1}-\frac{1}{x+5}=0.8$$ $$4=0.8(x+1)(x+5)$$ $$4x^2+24x=0$$ $$4x(x+6)=0$$

Solution: $x_1=0$ and $x_2=-6$

Definately very elegent! :)

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    $\begingroup$ Hint: $\frac{1}{(x+k)(x+k+1)} = \frac{1}{x+k} - \frac{1}{x+k+1}$ $\endgroup$ – achille hui Jul 17 '18 at 13:37
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HINT

We have that

$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=$$

$$=\frac{1}{(x+1)}\color{red}{-\frac{1}{(x+2)}+\frac{1}{(x+2)}-\frac{1}{(x+3)}+\frac{1}{(x+3)}-\frac{1}{(x+4)}+\frac{1}{(x+4)}}-\frac{1}{(x+5)}$$

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Make the equations more symmetric by shifting $z:=x+3$ and simplify

$$\frac{1}{(z-2)(z-1)}+\frac{1}{(z-1)z}+\frac{1}{z(z+1)}+\frac{1}{(z+1)(z+2)}$$

$$=\frac{2z^2+4}{(z^2-4)(z^2-1)}+\frac2{z^2-1}$$

$$=\frac{4z^2-4}{(z^2-4)(z^2-1)}$$

$$=\frac4{z^2-4}$$

giving $z=\pm 3$ and $x=0,x=-6$.

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  • $\begingroup$ Strange... I got a bit of a different result.. Am I wrong in my solutions? (I've edited my question) $\endgroup$ – L0wRider Jul 17 '18 at 13:57
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    $\begingroup$ @Maxim: there was a typo in the last result. $\endgroup$ – Yves Daoust Jul 17 '18 at 14:00
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Hint: your equation is equivalent to $$\frac{4x(6+x)}{5(1+x)(5+x)}=0$$

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With factors following an arithmetic progression, you establish the rule

$$\frac1{a(a+b)}+\frac1{(a+b)(a+2b)}=\frac{2a+2b}{a(a+b)(a+2b)}=\frac2{a(a+2b)}.$$

Applying it three times, the sum reduces to

$$\frac4{(x+1)(x+5)},$$ giving the easy quadratic equation

$$(x+1)(x+5)=\frac4{0.8}$$

or $$x(x+6)=0.$$

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