Could you please prove the following equation from Arithmetic Progressions?

If #a_1#, ... , #a_n# are in AP where, #a_i > 0# for all #i#, then show that
#1/(sqrt(a_1)+sqrt(a_2))+1/(sqrt(a_2)+sqrt(a_3))+....+1/(sqrt(a_n-1)+sqrt(a_n))= (n-1)/(sqrt(a_1)+sqrt(a_n))#.

1 Answer
Feb 25, 2018

See below.

Explanation:

If the sequence #a_k# is an AP then

#a_(k+1) = a_k + Delta# so

#1/(sqrt(a_k)+sqrt(a_(k+1))) = (sqrt(a_(k+1))-sqrt(a_k))/(a_(k+1)-a_k) = (sqrt(a_(k+1))-sqrt(a_k))/Delta#

and then

#sum_(k=1)^(n-1)1/(sqrt(a_k)+sqrt(a_(k+1))) =1/Delta (sum_(k=1)^(n-1) sqrt(a_(k+1))-sum_(k=1)^(n-1) sqrt(a_k)) = 1/Delta(sqrt(a_n)-sqrt(a_1))#

but

#1/Delta(sqrt(a_n)-sqrt(a_1)) = 1/Delta(sqrt(a_n)-sqrt(a_1))((sqrt(a_n)+sqrt(a_1))/(sqrt(a_n)+sqrt(a_1))) = 1/Delta(a_n-a_1)/(sqrt(a_n)+sqrt(a_1)) = 1/Delta ((n-1)Delta)/(sqrt(a_n)+sqrt(a_1)) = (n-1)/(sqrt(a_n)+sqrt(a_1))#