State where the power series is centered. Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) \" (-1)"X" 6) :1: n m (1)rz+1x,,l 7) \"2:1: 4.. n \"1 2 8) ;n+1( x) no x3n+1 Write an equivalent series with the index of summation beginning at n =1. Show that the function represented by the power series is a solution of the differential equation. so 2n+1 x 12 = \" =0 \" ;(2n+1)!'y y Find a geometric power series for the function centered at 0. 3 4+}: 14) f(x) = Find a power series for the function, centered at c, and determine the interval of convergence. 1 15) f(x)=1_3x,c=0 2 16) f(x)=2x+3, 6:0 1 \"c n ,, Use the power series 1 : Z(1) x to determine a power series, centered at 0, for the function. + x ":0 Identify the interval of convergence. 2 1 1 2 + x 1 1+x lx 18)f(x)= 1 =i[L] (\"1)2 dx x+1 17) h(x) =