24easy dude
r=1Σn4r-14r
=>r=1Σn1-14r
=>n-[1/4(1-1/4n)1-1/4]
=>n-4n-13.4n
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
24
23the 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]
1option c
n - 14n x 3 + 13
or n + 4n-13 x 4n closest i think but certainly not correct.
1its not my solution ,i am talking about the option c being closest to correct ans, which is post 6.
23kk.....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
