Discrete Mathematics with Graph Theory (3rd Edition) 156

Discrete Mathematics with Graph Theory (3rd Edition) 156 -...

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154 Solutions to Exercises Therefore, f(x) + 10xf(x) + 25x 2 = ao + (al + lOao)x + (a2 + lOal + 25ao)x2 + . .. + (an + lOan-l + 25an_2)X n + . ... So f(x)(l + lOx + 2 ) = 1 + 35x. Therefore, 1 = (1 + 5X)2 (1 + 35x) = + 2( -5x) + . .. + (n + 1)( -5xt + . .. )(1 + = 1 + 25x - 275x 2 + . .. + [(n + 1)( -5t + n( -5t- 1 (35)]xn + . = 1 + - 2 + . .. + [(1 - 6n)( _5)n]xn + . .. and so an = (1 - 6n)( _5)n. 11. [BB] + alX + a2x2 + a3x3 + + anx n + + 2a n _lX n + + an_2 Xn + + 2an_3 + 2xf(x) 2aox + 2alX2 + 2a2x3 + x 2 aox 2 + alx 3 + 2x 3 f(x 2aox3 + Therefore, - - x 2 + 3 = + (al - 2ao)x + - 2al - ao)x 2 + (a3 - 2a2 - al + 2ao)x3 + . .. + (an - 2an -l - an-2 + 2an_3)X n + . .. = 1 + x - x 2 since = 1, al - 2ao = 3 - 2(1) = 1, a2 = 6 - 2(3) - 1 = -1 and an - 2an -l - + 2an-3 = 0 for n ~ 3. So and f 1 + x -
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