Answer:
The correct option is:
2, 4, 12, 48, 240
Step-by-step explanation:
We have to determine the first 5 terms of the sequence by using following formula:
To Calculate the first 5 terms.
Put n = 1
a₁ = 2 (1)!
a₁ = 2(1)
a₁ = 2
Put n = 2
a₂ = 2 (2)!
a₂ = 2 (2·1) = 2(2)
a₂ = 4
Put n = 3
a₃ = 2(3)!
a₃ = 2(3·2·1)
a₃ = 2(6)
a₃ = 12
Put n = 4
a₄ = 2(4)!
a₄ = 2(4·3·2·1)
a₄ = 2(24)
a₄ = 48
Put n =5
a₅ = 2(5)!
a₅ = 2(5·4·3·2·1)
a₅ = 2(120)
a₅ = 240
Answer:
2x - y - 3z = 0
Step-by-step explanation:
Since the set
{i, j} = {(1,0), (0,1)}
is a base in
and F is linear, then
<em>{F(1,0), F(0,1)} </em>
would be a base of the plane generated by F.
F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k
F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k
Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k
We need a normal vector which is the cross product of 3i+2k and 4i-j+3k
(3i+2k)X(4i-j+3k) = 2i-j-3k
The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by
2(x-3) -1(y-0) -3(z-2) = 0
or what is the same
2x - y - 3z = 0