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