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

HINT

We have that

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

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


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 $x=0,x=-6$.


Hint: your equation is equivalent to $$\frac{4x(6+x)}{5(1+x)(5+x)}=0$$

Tags:

Polynomials

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