How do we approach this?

Algebra Level 5

The value of $\displaystyle \sum_{k=1}^{1000} \dfrac{k}{k^4+k^2+1}$

can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are coprime positive integers. Find $m+n$ .

1 solution

Krishna Sharma
Oct 6, 2014

Sorry for my wrong solution I just forgot to multiply by $\frac {1}{2}$

Here

$\displaystyle k^{4} + k^{2} + 1 = (k^{2} -k +1)(k^{2} + k +1)$

Rewriting it we get

$\displaystyle \sum_{k=1}^{1000} \frac {1}{2} \big (\frac {1}{k^{2} - k + 1} - \frac {1}{k^{2} + k + 1})$

When we will put the values of k all the terms will cancel out except first and last terms.

Finally we get

$\displaystyle \frac{1}{2} (1 - \frac {1}{1001001}) = \frac{500500}{1001001}$

Hence

$\displaystyle 500500 + 1001001 = \boxed{1501501}$

I thought a lot if 500500 and 1001001 are coprime or not ?!!! :)

Sandeep Bhardwaj - 6 years, 8 months ago

It can be easily seen that $1001001=2\cdot 500500 + 1$ so they're of course coprime.

Jubayer Nirjhor - 6 years, 8 months ago

Factorising 1001001 = $3 \times 333667$

And

500500 = $2^{2} \times 5^{3} \times 7 \times 11 \times 13$

Hence they are co-prime

Krishna Sharma - 6 years, 8 months ago

Yeah..I did the same way !

Sandeep Bhardwaj - 6 years, 8 months ago

$1001001-2\cdot 500500=1001001-1001000=1$ and hence they are coprime.

Rahul Saha - 6 years, 8 months ago

Please note that (k+1)^2-(k+1)+1 is actually equal to k^2+k+1 :)

Hoo Zhi Yee - 6 years, 7 months ago

guys pls suggest me some books to improve logical reasoning

Abhay Pandey - 6 years, 8 months ago

This is exactly what I did!

Eric LeClair - 6 years, 8 months ago

Dude, how do you factorise the k/(k^4+k^2+1)?

Shreyas Radhakrishna - 6 years, 8 months ago

write k^4+k^2+1 = (k^2 + 1)^2 - (k^2) which is a difference of two squares

Andre Chan - 6 years, 7 months ago