Find the value of n such that an+1+bn+1/an+bn may be geometric mean between a and b .
-
UP 0 DOWN 0 0 4
4 Answers
62(an+1+bn+1)/(an+bn) = √ab
=((a/b)n+1+1)/((a/b)n+1) = √a/b
=((a/b)n+1+1)= ((a/b)n+1)x√a/b
=(a/b)n+1+1= (a/b)n+1/2+(a/b)1/2
(a/b)n+1 - (a/b)n+1/2 = (a/b)1/2-1
(a/b)n+1/2 {(a/b)1/2 -1} = (a/b)1/2-1
{(a/b)n+1/2-1} {(a/b)1/2 -1} = 0
now did you get it?
only when n=-1/2
62divide the numerator by bn+1 and the denominator by bn
simultaneously divide the RHS by b