已知f(n)=n^2(n为正奇数时）f(n)= -n^2(n为正偶数） 若an=f(n)+f(n+1),求Sn

a1=f(1)+f(2)
a2=f(2)+f(3)
a3=f(3)+f(4)
..
an=f(n)+f(n+1)
Sn=f(1)+f(n+1)+2*[(f(n)+f(n-1))+(f(n-2)+f(n-3))+..+(f(3)+f(2))]
n奇数 f(n)+f(n-1)=n^2-(n-1)^2=2n-1
Sn=1-(n+1)^2+2*[(2n-1)+(2n-3)+...+5]=1-(n+1)^2+(2n-1+5)*(2n-1-5)/2=1-(n+1)^2+(2n+4)(n-3)
=1-(n+1)^2+(2n^2-2n-12)
n偶数 f(n)+f(n-1)=-n^2+(n-1)^2=1-2n
Sn=1+(n+1)^2+2*[(1-2n)+(3-2n)+..+5]
=1+(n+1)^2+(6-2n)(-n-2)
=1+(n+1)^2+(2n^2-2n-12)

a1 = f(1) + f(2) = 1² - 2²
a2 = f(2) + f(3) = -2² + 3²
...
an = f(n) + f(n+1)
Sn = f(1) + f(2) + f(2) + f(3) + f(3) + f(4) + ... f(n) + f(n+1)
= f(1) + f(2) + f...

n为奇数
an=f(n)+f(n+1)=n^2-(n+1)^2=-2n-1
n为偶数
an=f(n)+f(n+1)=-n^2+(n+1)^2=2n+1
n为奇数
an+a(n+1)=-2n-1+2(n+1)+1=2
n为偶数
sn=(a1+a2)+(a3+a4)+...+(a(n-1)+an)=n
n为奇数
sn=(a1+a2)+(a3+a4)+...+(a(n-2)+a(n-1))+an=n-1-2n-1=-n-2

an=f(n)+f(n+1)
Sn=f(1)+f(n+1)+2*[(f(n)+f(n-1))+(f(n-2)+f(n-3))+..+(f(3)+f(2))]
n奇数 f(n)+f(n-1)=n^2-(n-1)^2=2n-1
Sn=1-(n+1)^2+2*[(2n-1)+(2n-3)+...+5]=1-(n+1)^2+(2n-1+5)*(2n-1-5)/2=1-(n+1)^...