`(2^(n+4)-2xx2^(-n))/(2xx2^((n+3)))+2^(-3)` is equal to `2^(n+1)` b. `(9/8-2^n)` c. `(-2^(n+1)+1/8)` d. `1`

(2^(n+4)-2xx2^(n))/(2^(2)xx2^(n+3))+2^(-3) is equal to a.2^(n+1)b .((9)/(8)-2^(n))c*(-2^(n+1)+(1)/(8))d.1

(3xx9^(n+1)-9xx3^(2n))/(3xx3^(2n+3)-9^(n+1))

If 2^(n+4)-2^(n+2)=3, then n is equal to a.0 b.2c.-2d.-1

Simiplify : (16xx2^(n+1)-4xx2^(n))/(16xx2^(n+2)-2xx2^(n+1))

Simiplify : (16xx2^(n+1)-4xx2^(n))/(16xx2^(n+2)-2xx2^(n+1))

Simplify: frac(16 xx 2^(n+1)-8 xx 2^(n))(16 xx 2^(n+2)-4 xx 2^(n+1))

Sum of the first n terms of the series (1)/(2)+(3)/(4)+(7)/(8)+(15)/(16)+... is equal to 2^(n)-n-1 b.1-2^(-n) c.n+2^(-n)-1 d.2^(n)+1