(3+1)(3^(2)+1)(3^(4)+1)(3^(8)+1)(3^(16)+...

`(3+1)(3^(2)+1)(3^(4)+1)(3^(8)+1)(3^(16)+1)=?`

A

`((3^(32)-1))/(2)`

B

`((3^(16)-1))/(2)`

C

`((3^(64)-1))/(2)`

D

`((3^(128)-1))/(2)`

Verified by Experts

The correct Answer is:

A

(3+1)(3^(2)+1)(3^(4)+1)(3^(8)+1)(3^(16)+1)=?
Text Solution
|
Show that (1+3^(-1))(1+3^(-2))(1+3^(-4))(1+3^(-8)).......(1+3^(-2^(n))...
Text Solution
|
Find the sum 1-(1/8)+(1/8)*(3/16)-((1*3*5)/(8*16*24))+......
Text Solution
|
{(1/3)^2}^4 is equal to (a)(1/3)^6 (b) (1/3)^8 (c) (1/3)^(24) ...
Text Solution
|
1+(3+1)(3^2+1)(3^4+1)(3^8+1)(3^(16)+1)(3^(32)+1) is equal to (3^(64)-1...
Text Solution
|
(7)/(8)+(3)/(4)((1)/(2))(2)/(3)-(1)/(3))
Text Solution
|
(1-(2)/(1!)+(4)/(2!)-(8)/(3!)+....)(1+(4)/(1!)+(16)/(2!)+(64)/(3!)+......
Text Solution
|
(1)/(2)+(3)/(2).(1)/(4)+(5)/(3).(1)/(8)+(7)/(4).(1)/(16)+....=
Text Solution
|
-(2)/(3)-(1)/(2)((4)/(9))-(1)/(3)((8)/(27))-(1)/(4)((16)/(81))-…..oo=
Text Solution
|