1, 3, 6, 10, 15 ..... Let nth term = an^2 + bn + c Put n = 1 a + b + c = 1 4a + 2b + c = 3 9a + 3b + c = 6 5a + b = 3 3a + b = 2 2a = 1 a= 1/2 b = 2-3/2= 1/2 C = 0 So (1/2)n^2 + 1/2(n) Nth term = 1/1/2(n^2+n)= 2/n(n+1) => 2 [ 1/n - 1/(n+1)] 2 [ 1/1-1/2+1/2-1/3+.....+1/9-1/10] 2[ 1-1/10] 2 * 9/10 = 9/5