I. 3*2^0
II. 3*2^1
III 3*2^2
IV. 3*2^3
V. 3*2^4
VI. 3*2^5
The sum 3*(2^0+2^1+2^2+2^3+2^4+2^5)=3*(2^6-1)=3*(64-1)=3*63=189
D is the answer
Answer:
Step-by-step explanation:
Answer:

Step-by-step explanation:
We are given with two equations
first equation is 
second equation is

we find the result of subtracting two equation
subtract the second equation from the first, so
first equation - second equation, multiply second equation by -1 and then add it with first equation


Now add both equations, we get
