Written as unreduced radicals and arranged in order, the problem is:
1/(√16 + √14) + 1/(√14 + √12)+1/(√12 + √10)+ 1/((√10 + √8) * 1/(√8 + √6) + 1/(√6 + √14)
Now multiply each fraction by (its conjugate divided by its conjugate):
1/(√16 + √14)·(√16 - √14)/(√16 - √14) = (√16 - √14)/(16 - 14) = (√16 - √14)/2
1/(√14 + √12)·(√14 - √12)/(√14 - √12) = (√14 - √12)/(14 - 12) = (√14 - √12)/2
1/(√12 + √10)·(√12 - √10)/(√12 - √10) = (√12 - √10)/(12 - 10) = (√12 - √10)/2
...
1/(√6 + √4)·(√6 - √4)/(√6 - √4) = (√6 - √4)/(6 - 4) = (√6 - √4)/2
Note that all the denominators are 2, so the numerators can be added.
Now note that when you add all the numerators together, all the intermediate terms cancel and the only terms left in the numerator is the first and the last: √16 - √4 = 4 - 2 = 2
So this sumes to 2/2 = 1 <----- A quite surprising result! And, a super neat problem!
Written as unreduced radicals and arranged in order, the problem is:
1/(√16 + √14) + 1/(√14 + √12)+1/(√12 + √10)+ 1/((√10 + √8) * 1/(√8 + √6) + 1/(√6 + √14)
Now multiply each fraction by (its conjugate divided by its conjugate):
1/(√16 + √14)·(√16 - √14)/(√16 - √14) = (√16 - √14)/(16 - 14) = (√16 - √14)/2
1/(√14 + √12)·(√14 - √12)/(√14 - √12) = (√14 - √12)/(14 - 12) = (√14 - √12)/2
1/(√12 + √10)·(√12 - √10)/(√12 - √10) = (√12 - √10)/(12 - 10) = (√12 - √10)/2
...
1/(√6 + √4)·(√6 - √4)/(√6 - √4) = (√6 - √4)/(6 - 4) = (√6 - √4)/2
Note that all the denominators are 2, so the numerators can be added.
Now note that when you add all the numerators together, all the intermediate terms cancel and the only terms left in the numerator is the first and the last: √16 - √4 = 4 - 2 = 2
So this sumes to 2/2 = 1 <----- A quite surprising result! And, a super neat problem!