Let s = 2 + 3 + 6 + 11 + 18 + …………. + n
By shifting each term by one
S = 2 + 3 + 6 + 11 + 18 + ……… + nth ……… (1)
S = 2 + 3 + 6 + 11 + 18 + ……… + (n - 1)th + nth …(2)
by (1) - (2) we get
0 = 2 + 1 + 3 + 5 + 7 + …….nth - (n - 1)th - nth
Nth = 2 + (1 + 3 + 5 + 7 + 9 + …….2r - 1) ……….(3)
Nth = 2 + (summation of first (n - 1)th term)
We know by property that:
