(In reply to

Answer by K Sengupta)

Let the ith term of the given sequence be denoted by T(i).

Let us define:

D(i) = T(i+1) - T(i), and:

D_2(i) = D(i+1) - D(i)

Accordingly, we set up the following table:

T(i) D(i) D_2(i)

1

1

2 1

2

4 1

3

7 3

6

13 4

10

23 6

16

39 9

25

64 10

35

99 16

51

150 15

66

216

From the above table, we observe that:

D_2(1) = 1, D_2(3) = 3, D_2(5) = 6, D_2(7) = 10, D_2(9) = 15

Hence, the odd numbered terms of D_2(i) are triangular numbrs, so that:

D_2(11) = 21

Again, D_2(2) = 1, D_2(4) = 4, D_2(6) = 9, D_2(8) = 16

Hence, the even numbered terms of D_2(i) are square numbrs, so that:

D_2(10) = 25, and D_2(12) = 36

Accordingly, we must have:

D(11) = D(10) + D_2(10) = 66+25 = 91

-> T(12) = T(11) + D(11) = 216+91 = 307

D(12) = D(11) + D_2(11) = 91+21= 112

-> T(13) = T(12) + D(12) = 307+112= 419

D(13 = D(12) + D_2(12) = 112+36= 148

-> T(14) = T(13) + D(13) = 419+148= 567