Let {an} be an arithmetic sequence with the first term of 3 and the tolerance of D, and {BN} be a sequence determined by BN = an + an + 1, then what are the first n terms of {BN} and Sn?

an=3+(n-1)d
a(n+1)=3+nd
So BN = 6 + (2n-1) d = (6-D) + 2DN
So BN is an arithmetic sequence
b1=6-d+2d=6+d
So Sn = (B1 + BN) n / 2 = (12 + 2DN) n / 2 = DN & sup2; + 6N

If the sequence {an} is an arithmetic sequence and an = an ^ 2 + N, then the real number a=
2. In the arithmetic sequence with tolerance D, A6 + A10 = 18, then a1 + 7d=

An = a * n ^ 2 + Na1 = a + 1A2 = a * 4 + 2A3 = a * 9 + 3a2-a1 = a3-a2 = D4A + 2-a-1 = 9A + 3-4a-23a + 1 = 5A + 1A = 02 A6 + 2D = a8a8 + 2D = a10a6 + A10 = a8-2d + A8 + 2D = 2a8 = 18a8 = 9a1 + 7

The sequence {an} satisfies the recursive formula an = 3A (n-1) + 3 ^ n-1 (n > = 2), and A1 = 5 makes {(an + y) / 3 ^ n} the real number y of arithmetic sequence=

Let BN = (an + y) / 3 ^ n be an arithmetic sequence, then bn-b (n-1) is a constant, bn-b (n-1) = (an + y) / 3 ^ n - [a (n-1) + y] / 3 ^ (n-1), and then substitute an = 3A (n-1) + 3 ^ n-1 to obtain bn-b (n-1) = 1 - (1 + 2Y) / 3 ^ NY, which is a real number and cannot be an algebraic expression of N, so 1 + 2Y = 0y = - 1 / 2

Given that the sum of the first n terms of the sequence {an} is Sn = a ^ n-2 (a is a real number that is not zero), then is the sequence {an} equal ratio or equal difference?

1 find the general formula of an 2 whether there are three terms AR, as, at (r less than s less than t) in this sequence to be equal difference an + 2 is equal ratio sequence. An + 2 = (a1 + 2) 2 ^ (n-1) = 2 ^ (n + 1) an = 2 ^ (n + 1)

Let {an} be an arithmetic sequence with the first term of 3 and the tolerance of D, and {BN} be a sequence determined by BN = an + an + 1, then what are the first n terms of {BN} and Sn?