( 3^2n ) - 1=k2^(n+2)
or (9^n)-1=4*2^n
or (9^n)=4*2^n+1

so for n=1,LHR=RHS

Ans option A

What's the OA ?

Aman verma wrote:Q: ( 3^2n ) - 1 is divisible by 2^(n+3) for n =
a)1 b)2 c)3 d)4 e)None of these

(3^2n) - 1 / 2^(n+3)

IMO E

Aman verma wrote: It's 2^(n+3). Now corrected !

Now it should be
( 3^2n ) - 1=k2^(n+3)
or (9^n)-1=k8*2^n
or (9^n)=k8*2^n+1

For n=1 ,(9^n)=9..or 8=k*8*2^n..this is possible for n=0 and k=1..not possible
For n=2 ,(9^n)=81..or 80=k*8*4..or k=5/2..not possible
For n=3 ,(9^n)=729..or 728=k*8*8..or k=81/8..not possible
For n=4 ,(9^n)=6561..or 6560=k*8*16..or k=205/4..not possible

Ans E

OAE. Thanks to all . Regret the error !

Hello - thanks for the post and answer. I'm having troubles understanding the alegbra. Can someone help explain how k2^(n+3) becomes k8*2^n+1 ?


i would appreciate the help. thanks

k2^(n+3)=K8*2^n

principle is a^b * a^c = a^(b+c)

k2^n * 2^3=8k2^n

1 is added to the right side since it was on the left side

hope that helpbs

thanks. I think i get it.. but did you mean: k2^n * 2^3=8k2^n to read k2^n * 2k^3=8k2^n


in other words, was the K missing in your post above?

ohhh nevermind I get it..thanks again.

cheers

ohhh nevermind I get it..thanks again.

cheers