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

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

step-by-step explanation:

two pairs of adjacent sides are of equal length. this implies that one diagonal divides the kite into congruent triangles, and so the angles between the two pairs of equal sides are equal in measure. it also implies that the diagonals are perpendicular.