高一数列求和题1.等比数列的首项为a,公比为q,Sn为前n项的和,求S1+S2+…+Sn2.数列{an}的通项公式是an=1/[√n+√(n+1）],若前n项和为10,则项数n为?

1、Sn=a(q^n-1)/(q-1)
S1+S2+…+Sn=a/(q-1)[q(q^n-1)/(q-1)-n]=a[q^(n+1)-(n+1)q+n]/(q-1)^2
2、an=1/[√n+√(n+1)]=√(n+1)-√n
Sn=√(n+1)-1=10
n=120

1.
an=aq^(n-1)
Sn=a(q^n-1)/(q-1)
S1+S2+…+Sn
=[a(q^1-1)/(q-1)]+[a(q^2-1)/(q-1)]+[a(q^3-1)/(q-1)]+……+{a[q^(n-1)-1]/(q-1)}+[a(q^n-1)/(q-1)]
=[a/(q-1)]{(q^1-1)+(q^2-1)+(q^3-1)+……+[q^(n-1)-1]+(q^n-1)}
=[a/(q-1)][q^1+q^2+q^3+……+q^(n-1)+q^n-n]
=[a/(q-1)][q(q^n-1)/(q-1)-n]
=aq(q^n-1)/(q-1)^2-na/(q-1)
2.
an=1/[√n+√(n+1)]
=[-√n+√(n+1)]/{[√n+√(n+1)][-√n+√(n+1)]}
=-√n+√(n+1)
an=-√n+√(n+1)
a(n-1)=-√(n-1)+√n
a(n-2)=-√(n-2)+√(n-1)
……
a3=-√3+√4
a2=-√2+√3
a1=-√1+√2
两加相加：
sn=√(n+1)-1
sn=√(n+1)-1=10
√(n+1)=11
n+1=121
n=120