已知s(2n+1)=4n^2+2n,求an

s(2n+1)=4n^2+2n
S(2n + 1) = (2n + 1 -1)^2 + 2n + 1 -1
S(n) = (n-1)^2 + n - 1 = n^2 -n

Sn = a1 + a2 + …… + an
S<n-1> = a1 + a2 + …… + a<n-1>

所以
an = Sn - S<n-1>
= n^2 - n - [(n-1)^2 - (n-1)]
= n^2 - n - n^2 + 2n - 1 + n - 1
= 2n -2

因为S(2n+1) = 4n^2+4n+1-1-2n = (2n+1)^2-(1+2n)

（说明：不用配方就用代换思想，设t=2n+1,n=(t-1)/2,代入s(2n+1)=4n^2+2n,得
S(n) = n^2-n ）

所以，S(n) = n^2-n , S(n+1) = (n+1)^2-(n+1)

an = S(n+1)-S(n) = 2n