Find the solution of the recurrence relation an = 3an−1 −3an−2 +an−3 if a0 = 2, a1 = 2, and a2 = 4.

Using the recurrence relation, we can find a couple more values in the sequence:

First differences are 0, 2, 4, 6, ...

Second differences are constant at 2, so the function is quadratic.

The sequence can be described by the quadratic ...

... an = n² -n +2

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We know the value for n=0 is 2, so we can find a and b using the given values for a1 and a2.

... an = an² +bn +2

... a1 = 2 = a·1² +b·1 +2 . . . . for n=1

... a + b = 0

... a2 = 4 = a·2² -a·2 +2 . . . . for n = 2; using b=-a from the previous equation

... 2 = 2a

... a = 1 . . . . so b = -1

the answer is 14ft

step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}