0
$\begingroup$

can we compute this term without using calculator?

$$\frac{1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{997^2}+\frac{1}{999^2}-\frac{1}{1002^2}-\frac{1}{1004^2}-\frac{1}{1006^2}-...-\frac{1}{1998^2}-\frac{1}{2000^2}}{1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}}$$

I can't see any pattern.

I tried to add and substract as mentioned in the comments but can't see how it helps.

thanks.

$\endgroup$
  • 1
    $\begingroup$ In the numerator add and substrate the even terms present in the denominator (1/2^2 +1/4^2+... ). $\endgroup$ – incognito Oct 27 '15 at 15:09
  • $\begingroup$ @incognito i tried that now but can't see how it helps. $\endgroup$ – tammy Oct 27 '15 at 16:28
  • $\begingroup$ @tammy: Now factor $4$ in the denominator of each. $\endgroup$ – Lucian Oct 28 '15 at 1:43
1
$\begingroup$

The given fraction $$\frac{1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{997^2}+\frac{1}{999^2}-\frac{1}{1002^2}-\frac{1}{1004^2}-\frac{1}{1006^2}-...-\frac{1}{1998^2}-\frac{1}{2000^2}}{1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}}$$ can be rewritten as $$\frac{\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}\right)-\left(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{1000^2}+\frac{1}{1002^2}+...+\frac{1}{1998^2}+\frac{1}{2000^2}\right)}{1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}}$$ by adding and subtracting $\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{1000^2}$ to the numerator of the fraction. The above fraction can now be written as: $$1 - \frac{\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{1000^2}+\frac{1}{1002^2}+\frac{1}{1004^2}+...+\frac{1}{1998^2}+\frac{1}{2000^2}}{1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}}$$

$$ = 1 - \frac{\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}\right)}{1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{999^2}+\frac{1}{1000^2}} = 1 - \frac{1}{2^2} = \frac{3}{4}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.