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3 Answers

Best answer
38 votes
38 votes

$S(n) = 3 + 6x^2 + 9x^4 + 12x^6 + ..$
$-x^2S(n) =\quad\; - 3x^2 - 6x^4 - 9x^6 - ...$
$(1-x^2)S(n) = 3 + 3x^2 + 3x^4 + 3x^6 + ...$
$(1-x^2)S(n) = 3 (1 + x^2 + x^4 + x^6 + ...)$

$(1-x^2)S(n) = 3 \times \frac{1}{(1-x^2)}$                   this series is a GP having infinite number and it a = 1 and r = x (i.e < 1)
$S(n) = \frac{3}{(1-x^2)^2}$ 
 

 So Answer will be C

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12 votes
12 votes

ans C

i solved it in another way

Given |x| <1, let x= 1/2

3+6 x 2 + 9 x 4 +12 x 6... = 3(1+2 x 2 + 3 x 4+4x 6) =3(1+ 2*1/4+3*1/16+4*1/64) = 21/4  ~ 5

Check in each option....

a) if x=1/2 , 3/(1+x 2)=12/5                             <5  Not the answer

b)  if x=1/2 , 3/(1+x 2)2 = 48/25                      <5  Not the answer

c)  if x=1/2 , 3/(1-x 2)2=16/3                             ~ 5  Hence the answer

d)  if x=1/2 , 3/(1-x 2)=4                                    <5  Not the answer

2 votes
2 votes

sum of infinite ap-gp is a/(1-r) + d*r/(1-r2 )

so,   3/1-x2 + 3x2 /(1-x2 )

so,3/(1-x2 )2

Answer:

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