Post by vsiap on Nov 16, 2014 4:07:15 GMT
The characteristic polynomial equals x3 −6x2 +12x−8 = (x−2)3. We know from Theorem 4 on page 519 that an = c12n + c2n2n + c3n22n for some constants c1, c2, c3. By substituting n = 0, 1, 2 into this equation, we obtain the linear system
The solution is c1 = −5 , c2 = 1/2, c3 = 13/2, and therefore
c1 =−5, 2c1 +2c2 +2c3 =4 and 4c1 +8c2 +16c3 =88.
The solution is c1 = −5 , c2 = 1/2, c3 = 13/2, and therefore
an = −5·2n + n·2n−1 + 13·n2 ·2n−1 for n = 0,1,2,3,....