a_1=-2, a_n=2(a_n-1)^2,a_4
a_n=ln(e^n+2), a_5
b_0=1, b_1=2, b_n+1=2b_n-b_0
Steve
answered
a1 = -2
a2 = 2(a1)^2 = 2(-2)^2 = 8
a3 = 2(a2)^2 = 2(8)^2 = 128
...
a1 = ln(e^2) = 2
a2 = ln(e^4) = 4
...
b0 = 1
b1 = 2
b2 = 2(2)-1 = 3
b3 = 2(3)-1 = 5
...
Explain Bot
answered
For sequence a:
a_1 = -2
a_n = 2(a_n-1)^2
To find a_4, we substitute n=4 into the recurrence relation:
a_4 = 2(a_4-1)^2
a_4 = 2(a_3)^2
To find a_3, we need to continue substituting:
a_3 = 2(a_2)^2
a_3 = 2(2(a_1)^2)^2
a_3 = 2(2(-2)^2)^2
a_3 = 2(8)^2
a_3 = 2(64)
a_3 = 128
Now that we have a_3, we can substitute it back into the equation to find a_4:
a_4 = 2(a_3)^2
a_4 = 2(128)^2
a_4 = 2(16384)
a_4 = 32768
Therefore, a_4 = 32768.
Now let's move on to sequence b:
b_0 = 1
b_1 = 2
b_n+1 = 2b_n - b_0
To find b_2, we substitute n=1 into the recurrence relation:
b_2 = 2b_1 - b_0
b_2 = 2(2) - 1
b_2 = 3
Similarly, to find b_3:
b_3 = 2b_2 - b_1
b_3 = 2(3) - 2
b_3 = 4
Continuing the pattern, we can find b_4:
b_4 = 2b_3 - b_2
b_4 = 2(4) - 3
b_4 = 5
Finally, to find b_5:
b_5 = 2b_4 - b_3
b_5 = 2(5) - 4
b_5 = 6
Therefore, b_5 = 6.
