(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