Question

1/(x-1)x+1/x(x+1)+...+1/(x+9)(x+10)=11/12

Answer 1

Notice that

`\frac1{(x-1)x} = \frac1{x-1} - \frac1x \\\\ \frac1{x(x+1)} = \frac1x - \frac1{x+1} \\\\ \vdots \\\\ \frac1{(x+9)(x+10)} = \frac1{x+9} - \frac1{x+10}`

That is, the n-th term (where n = -1, 0, 1, …, 9) in the sum on the left side has a partial fraction decomposition of

`(1)/((x+n)(x+n+1)) = \frac1{x+n} - \frac1{x+n+1}`

and in the sum, some adjacent terms will cancel and leave you with

`\frac1{(x-1)x} + \frac1{x(x+1)} + \cdots + \frac1{(x+9)(x+10)} = \frac1{x-1} - \frac1{x+10} = (11)/(12)`

Now solve for x, bearing in mind that we cannot have x = 0, -1, -2, …, -10 :

`\frac1{x-1} - \frac1{x+10} = (11)/(12)`

Combine the fractions on the left side:

`((x+10)-(x-1))/((x-1)(x+10)) = (11)/((x-1)(x+10)) = (11)/(12)`

Then we must have

`(x-1)(x+10) = x^2 + 9x - 10 = 12 \implies x^2+9x-22 = (x+11)(x-2) = 0`

so that either x = 2 or x = -11.