Unformatted text preview: + ... ; hence, an = (n + 1)2n is the solution. Our solution is exactly the same as the sequence in Problem 21 and so our answer here verities the result obtained earlier. 7. [BB] f(x) = xf(x) = ao + alX + a2x2 + + anx n + aox + alx 2 + ... + an_lX n + Subtracting gives f(x) xf(x) (1 x)f(x) ao + (al ao)x + ... + (an an_l)X n + ... 5 + 2x + 4x 2 + ... + 2 n x n + . . . (1 + 2x + 4x 2 + ... + 2 n x n + ... ) + 4 since ao = 5 and an anl = 2n. Thus, f(x) = _1_(1 + 2x +4x 2 + ... + 2 n x n + ... ) +4(_1_) Ix Ix = [1 + (2 + l)x + (4 + 2 + l)x 2 + ... + (2n + 2 n1 + 2 n2 + ... + 2 + l)xn + ... J + 4(1 + x + x 2 + ... + xn + ... ) = (1 + 3x + 7x 2 + ... + (2 n + 1 l)xn + ... ) + (4 + 4x + 4x 2 + ... + 4xn + ... ) Note that 1 + 2 + 22 + ... + 2 n = 2n2+~11 = 2 n + 1 1 as the sum of a geometric sequence. Thus, an = 2 n + 1 1 +4 = 2 n + 1 +3....
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 Summer '10
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 Graph Theory, Summation, Geometric progression, AO, Xn, 1  l

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