Solve $\left(\frac{1}{25\cdot26}+\frac{1}{26\cdot27}+\frac{1}{27\cdot28}+\frac{1}{28\cdot29}+\frac{1}{29\cdot30}\right)\cdot150+\frac{1}{x-1}=11$

Question

Solve for $x$: $$\left(\frac{1}{25\cdot26}+\frac{1}{26\cdot27}+\frac{1}{27\cdot28}+\frac{1}{28\cdot29}+\frac{1}{29\cdot30}\right)\cdot150+\frac{1}{x-1}=11$$

I have tried alot of things, but nothing seems to work. I imagine there is a really elegant solution that I just can't see. Please help, thanks!

Answer 1

You have

$$\frac 1{n(n+1)}=\frac 1n-\frac 1{n+1},$$

so you get a telescoping sum, which gives you

$$\left(\frac 1{25}-\frac 1{30}\right)\times 150+\frac 1{x-1}=11$$

so

$$\frac 5{5\times 150}\times 150 +\frac 1{x-1}=11$$

so

$$1+\frac 1{x-1}=11.$$

Finally, $\frac 1{x-1}=10$, so $x=\frac 1{10}+1=\frac {11}{10}$.

Answer 2

Hint: show that $$\left(\frac{1}{25\cdot 26}+...\frac{1}{29\cdot30}\right)\cdot 150=1$$