If #n/(n-1) * 1/n * n/(n+1) = 5/k# for positive integers #n# and #k#, what is the value of #k#?

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Answer 1 · 2018-03-19T08:08:38

#n=5# and #k=24#

#n/(n-1) * 1/n * n/(n+1) = 5/k#

#=>n/((n-1)(n+1))=5/k#

or #n/(n^2-1)=5/k#

or #5n^2-5=nk#

or #5n^2-nk-5=0#

Now discriminant is #k^2+100#

and #n=(k+-sqrt(k^2+100))/10#

As #n# and #k# are whole numbers, it is apparent that an important condition would be that #k^2# and #k^2+100# both are perfect squares.

Now in the series of square numbers, we have

#1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,441,484,529,576,625,676,729,....#

and it is only #576# and #676#, which are squares and have a difference of #100#.

Now, if #k^2=576#, #k=24# and #n=(24+-26)/10#

i.e. #n=5# or #-1/5#, but #n# and #k# are whole numbers

Hence, #n=5# and #k=24#