eure ..put n =1 in ur ans ...it isnt coming 3/4 ...
Find the sum to n terms of the series :
3/4 + 15/16 + 63/64 + .......=
(a) n-(4^n/3) + 1/3
(b) n-(4^n/3) - 1/3
(C) n-(4^-n/3) +1/3
(D) n-(4^-n/3) - 1/3
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13 Answers
qwerty
·2009-11-19 10:12:49
the question is
\sum{1 - \frac{1}{4^{n}}}
=n - \frac{\frac{1}{4}[ \frac{1}{4^{n}} - 1 ] }{\frac{1}{4}-1}
or simply put n = 1 [3]
Maths Musing
·2009-11-19 10:24:18
option c
n - 14n x 3 + 13
or n + 4n-13 x 4n closest i think but certainly not correct.
Maths Musing
·2009-11-19 10:33:50
its not my solution ,i am talking about the option c being closest to correct ans, which is post 6.
qwerty
·2009-11-19 10:35:33
kk.....but we cant mark that one if it comes in the exam ...lolz ...i think there is a typo by soumi while posting d ques