若等差数列{an}的前n项和为Sn,已知a(m-1)+a(m+1)+a(m²)=0,S(2m-1)=38,则m=
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若等差数列{an}的前n项和为Sn,已知a(m-1)+a(m+1)+a(m²)=0,S(2m-1)=38,则m=
若等差数列{an}的前n项和为Sn,已知a(m-1)+a(m+1)+a(m²)=0,S(2m-1)=38,则m=
若等差数列{an}的前n项和为Sn,已知a(m-1)+a(m+1)+a(m²)=0,S(2m-1)=38,则m=
条件有误,
把a(m-1)+a(m+1)+a(m²)=0改为a(m-1)+a(m+1)-a(m²)=0
S(2m-1)=(2m-1)am=38,显然am≠0,
由a(m-1)+a(m+1)-a(m²)=0得2am-a(m²)=0,
所以am=2,
所以(2m-1)am=38即2(2m-1)=38,m=10.
a(m-1)+a(m+1)+a(m2)=0
an=a1+(n-1)d
a1+(m-2)d+a1+md+a1+(m^2-1)d=0
2a1+2(m-1)d+(m+1)a1+(m+1)(m-1)d-ma1=0
2a(m)+(m+1)a(m)-ma1=0
(m+3)a(m)-ma1=0 (*)
a(m)=ma1/(m+3)
Sn=na1+n(n-1)d/2
(2m-1)a1+(2m-1)(2m-1-1)d/2=38
(2m-1)a1+(2m-1)(m-1)d=38
(2m-1)(a1+(m-1)d)=38
(2m-1)a(m)=38 (**)
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