Answer:
f) a[n] = -(-2)^n +2^n
g) a[n] = (1/2)((-2)^-n +2^-n)
Step-by-step explanation:
Both of these problems are solved in the same way. The characteristic equation comes from ...
a[n] -k²·a[n-2] = 0
Using a[n] = r^n, we have ...
r^n -k²r^(n-2) = 0
r^(n-2)(r² -k²) = 0
r² -k² = 0
r = ±k
a[n] = p·(-k)^n +q·k^n . . . . . . for some constants p and q
We find p and q from the initial conditions.
__
f) k² = 4, so k = 2.
a[0] = 0 = p + q
a[1] = 4 = -2p +2q
Dividing the second equation by 2 and adding the first, we have ...
2 = 2q
q = 1
p = -1
The solution is a[n] = -(-2)^n +2^n.
__
g) k² = 1/4, so k = 1/2.
a[0] = 1 = p + q
a[1] = 0 = -p/2 +q/2
Multiplying the first equation by 1/2 and adding the second, we get ...
1/2 = q
p = 1 -q = 1/2
Using k = 2^-1, we can write the solution as follows.
The solution is a[n] = (1/2)((-2)^-n +2^-n).
Answer:
Slope is 1/4
Step-by-step explanation:
No . because you would then have 2/4 which is 1/2 50%
Proof by induction
Base case:
n=1: 1*2*3=6 is obviously divisible by six.
Assumption: For every n>1 n(n+1)(n+2) is divisible by 6.
For n+1:
(n+1)(n+2)(n+3)=
(n(n+1)(n+2)+3(n+1)(n+2))
We have assumed that n(n+1)(n+2) is divisble by 6.
We now only need to prove that 3(n+1)(n+2) is divisible by 6.
If 3(n+1)(n+2) is divisible by 6, then (n+1)(n+2) must be divisible by 2.
The "cool" part about this proof.
Since n is a natural number greater than 1 we can say the following:
If n is an odd number, then n+1 is even, then n+1 is divisible by 2 thus (n+1)(n+2) is divisible by 2,so we have proved what we wanted.
If n is an even number" then n+2 is even, then n+1 is divisible by 2 thus (n+1)(n+2) is divisible by 2,so we have proved what we wanted.
Therefore by using the method of mathematical induction we proved that for every natural number n, n(n+1)(n+2) is divisible by 6. QED.
