General term, T_n = 1/( (2n-1)^(1/2)+(2n+1)^(1/2))Tn=1(2n−1)12+(2n+1)12
Rationalize the General term of the sequence :-
rArr T_n = ( (2n-1)^(1/2)-(2n+1)^(1/2))
/[( (2n-1)^(1/2)+(2n+1)^(1/2)).( (2n-1)^(1/2)-(2n+1)^(1/2))]⇒Tn=(2n−1)12−(2n+1)12((2n−1)12+(2n+1)12).((2n−1)12−(2n+1)12)
rArr T_n = ((2n-1)^(1/2)-(2n+1)^(1/2))/(-2)⇒Tn=(2n−1)12−(2n+1)12−2
rArr T_n = ((2n+1)^(1/2)-(2n-1)^(1/2))/2⇒Tn=(2n+1)12−(2n−1)122
Now summing all the terms ie.
rArr sum T_n = sum_(n=0)^n((2n+1)^(1/2)-(2n-1)^(1/2))/2⇒∑Tn=n∑n=0(2n+1)12−(2n−1)122
rArr sum T_n = [(3-1)+(5-3)+.............+((2n+1)^(1/2)-(2n-1)^(1/2))]/2
rArr sum T_n = [(cancel3-1)+(cancel5-cancel3)+.............+((2n+1)^(1/2)-cancel(2n-1)^(1/2))]/2...........................($)
:. sum T_n = [(2n+1)^(1/2)-1]/2....................option (4)
NOTE :- In equation ($) we should note a Pattern of
cancellation of the terms has formed ; where the initial
terms are eliminated by the succeeding terms at some point ,
except the Least and the Biggest terms which are 1 and
(2n+1)^(1/2) respectively and the term (2n-1)^(1/2) also gets
cancelled.
