Recitation 16 Now substituting a =0 and r=l gives the system of equations 1=A+B 1 A B Solving the system, we find that A=1/4 and B=3/4. Therefore, we have /43/4 1+x (d) Find a closed-form expression for Tn from the partial fractions expansion Solution. Using the formula for the sum of an infinite geometric series gives f(x)=(1-x+x2-x3+x4-…)+(1+3x+32x2+3x32+3x4+…) Thus, the coefficient of xn is: 3 (-1)+x·32Recitation 16 7 Now substituting x = 0 and x = 1 gives the system of equations: 1 = A + B 1 A B − =4 2 − 2 Solving the system, we find that A = 1/4 and B = 3/4. Therefore, we have: 1/4 3/4 f(x) = + 1 + x 1 − 3x (d) Find a closed­form expression for Tn from the partial fractions expansion. Solution. Using the formula for the sum of an infinite geometric series gives: 1 � � 3 � � 1 − x + x 2 3 4 2 3 4 4 f(x) = 4 − x + x − . . . + 4 1 + 3x + 3 x 2 + 3 x 3 + 3 x + . . . Thus, the coefficient of xn is: 1 3 3n Tn = 4 · (−1)n + 4 ·