等差数列中,a5+a10=-2,Sn是前n项和,S14=（）

1.∵等差数列{an}
∴S14=14（a1+a14）/2
a5+a10=a1+a14=-2
∴S14=-14
2.∵a10/a7=a1q^9/a1q^6=q^3=-1/27
∴q=-1/3
(a1+a3+a5+a7+a9)/(a2+a4+a6+a8+a10)
=a1(1+q^2+q^4+q^6+q^8)/a1(q+q^3+q^5+q^7+q^9)
=(1+q^2+q^4+q^6+q^8)/q(1+q^2+q^4+q^6+q^8)
=1/q=-3
3.∵f(0)=1/2 ∴a1=1/2
f(1)=a1+a2+a3+a4+...+an
∵f(1)=n^2an
f(1)=n^2an
∴Sn=n^2an
S[n]=n^2(S[n]-S[n-1])
n^2/(n^2-1)=S[n]/S[n-1]
n^2/(n-1)(n+1)=S[n]/S[n-1]
S[n]/S[n-1]*S[n-1]/S[n-2]*S[n-2]/S[n-3].**S[2]/S[1] =
n^2(n-1)^2(n-2)^2.*2^2/(n-1)(n+1)(n-2)(n)(n-1)(n-3)(n-2)(n-4)..*3*1
=n^2*2^2/(n+1)n*2*1
=2n/(n+1)
S[n]/S[1]=2n/(n+1)
∴s[n]=n/(n+1)

{1}根据等差数列的前N项和可得S14=14{a1+a14}/2
因为a1+a14=a5+a10这是等差数列的性质
所以可得到S14=14
{2}因为a10/a7=a1q的9次方/a1q的6次方=q立方=1/27
所以q=1/3
{a1+a3+a5+a7+a9}/{a2+a4+a6+a8+a10}
=a1{1+q2+q4+q6+q8}/...