Sequence {an} satisfies A1 = 1 / 2, a1 + A2 +.. + an = n square an, find an

A1 + A2 +.. + an = Sn = n & sup2; * an so when n > = 2, s (n-1) = (n-1) & sup2; * a (n-1) subtracts an = SN-S (n-1) = n & sup2; * an - (n-1) & sup2; * a (n-1) (n & sup2; - 1) * an = (n-1) & sup2; * a (n-1) n > = 2, that is, n-1 is not equal to 0, so n-1 is reduced to (n + 1) * an = (n-1) * a (n-1) an / a (n-1) =

It is known that the sum of the first n terms of the sequence {an} is Sn, and Sn = n-5an-85, n ∈ n*
Find the general term formula of sequence {Sn}, and find the minimum integer n that makes s (n + 1) > Sn hold

Sn = n-5an-85 (1) s (n + 1) = n + 1-5a (n + 1) - 85 (2) (2) - (1) we get 6A (n + 1) = 1 + 5An, that is, a (n + 1) - 1 = (5 / 6) (an-1) and A1 = - 14 from S1 = A1 = 1-5a1-85, so {an-1} is the first term-15, the equal ratio sequence of common ratio 5 / 6, so an = (- 15) * (5 / 6) ^ (n-1) + 1sn = (- 15) * [(5 / 6) ^ 0

The sum of the first n terms of an is Sn = 1-5 + 9-13 + 17-21 + +(- 1) ^ n-1 * (4n-3), find SN

The sum of the first n terms of sequence an is Sn = 1-5 + 9-13 + 17-21 + +When n is even, let n = 2K, K ∈ n *. Sn = s [2K] = (1-5) + (9-13) + (17-21) +... + {[4 (2k-1) - 3] - [4 (2k) - 3]} = - 4-4-4 -... - 4 [total K terms] = - 4K = - 2n. When n is odd, let

If the first n terms of sequence {an} and Sn = 1-5 + 9-13 + 17-21 +... + (- 1) ^ n-1 (4n-3) are known, then s22-s11=

S22=-4*11=-44
S11=1+4*5=21
S22-S11=-44-21=-65

Sequence {an} satisfies A1 = 1 / 2, a1 + A2 +.. + an = n square an, find an