Please find the sequence 18,34,55,81 , 1 / 2 [(n-3) (5N + 12)] + 18 Please find the sequence 18,34,55,81 , 1 / 2 [(n-3) (5N + 12)] + 18. The rule of this sequence is that the difference between them is an arithmetic sequence,

Please find the sequence 18,34,55,81 , 1 / 2 [(n-3) (5N + 12)] + 18 Please find the sequence 18,34,55,81 , 1 / 2 [(n-3) (5N + 12)] + 18. The rule of this sequence is that the difference between them is an arithmetic sequence,

General term reduction an = 1 / 2 [(n-3) (5N + 12)] + 18 = (5nn-3n) / 2
a1=(5*1*1-3*1)/2=1
a2=(5*2*2-3*2)/2=7
a3=(5*3*3-3*3)/2=18
a1+a2+a3+…… A (n-1) + an = sum of general terms = explain first
Sum of squares Formula 1 ^ 2 + 2 ^ 2 + +n^2=[(n+1)^3-3n(n+1)/2-(n+1)]/3
=(n+1)(n^2+2n+1-3n/2-1)/3
=(n+1)(2n+n)/6
=n(n+1)(2n+1)/6
The first n terms of natural sequence and formula 1 + 2 + 3 + + +n=n(n+1)/2
5nn=5n^2=5n(n+1)(2n+1)/6
3n =3n(n+1)/2
Substituting the sum of general terms = 5N (n + 1) (2n + 1) / 12-3n (n + 1) / 4
Simplify n (n + 1) (5n-2) / 6
If the sequence starts from the third term, A3 = 18;
Then the sum of the first n terms of the sequence is the sum of the general terms - a2-a1 = n (n + 1) (5n-2) / 6-8 (n starts from 3)
If n starts from 1
So B1 = A3 = 18
Then the sum of the first n terms = (n + 2) (n + 3) (5N + 8) / 6-8 (n starts from 1)