progression

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S[n]= 3/4 +15/16 +63/64+......
find S[n]

a) n - 1/3*4^(n) -1/3
b) n + 1/3*4^(-n) -1/3
c) n + 1/3*4^(n)- 1/3
d) n-1/3*4^(-n) +1/3

#Mathematics #Algebra

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7 Answers

1

Philip Calvert ·2009-11-26 09:14:38

This is simple stuff...

S(n) = n - (1/4 + 1/16 + 1/64....n terms)

= n - 1/4(1 - (1/4)ⁿ)3/4

= n - 1/3 + (1/3)4^-n...

so (b)

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1

Maths Musing ·2009-11-26 09:15:32

the correct is option c .

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1

archana anand ·2009-11-26 09:19:13

if wwe put n=1.....we should get3/4
so ans c.

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1

Philip Calvert ·2009-11-26 09:24:09

@archana : if we put n = 1 in (C) we get.. 1 + 4/3 - 1/3 which is 2

i think 1/3*4 means (1/3)4 and not 1/(3*4)...

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24

eureka123 ·2009-11-26 18:14:38

why are the ques being repeated here ?????

it was done hardly week back...http://targetiit.com/iit-jee-forum/posts/sum-of-series-12610.html

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1

archana anand ·2009-11-26 23:19:48

@philip okie i misunderstood it....the option is b then

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1

Soumi Dasgupta ·2009-11-29 04:23:54

Amio eta post korechi giri!!!

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