{an}是等差数列前n项和Sn已知Sm=a Sn-Sn-m=b 求Sn

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/23 20:59:10
{an}是等差数列前n项和Sn已知Sm=a Sn-Sn-m=b 求Sn
xRJ0.Bͺ) ns)Ae@\|"ſ&Mkƍ{#zAw?ꩻ| ?;7q^{)4p\N٢[i"Zum$jJʆIQa th#GjJv2Z"-sz!#墩h,w2y*67RH xW@ H")Ds>FG܃[*e^_;>dacv\^" ExmfU/ vˉ@g EW81 i&.Zu~u

{an}是等差数列前n项和Sn已知Sm=a Sn-Sn-m=b 求Sn
{an}是等差数列前n项和Sn已知Sm=a Sn-Sn-m=b 求Sn

{an}是等差数列前n项和Sn已知Sm=a Sn-Sn-m=b 求Sn
Sn-S(n-m)=A(n-m+1)+A(n-m+2)+……+A(n-m+m)=b 共m项
A(n-m+1)=A1+(n-m)d
A(n-m+2)=A2+(n-m)d
……
A(n-m+m)=An=Am+(n-m)d
Sn-S(n-m)
=A(n-m+1)+A(n-m+2)+……+A(n-m+m)
=(A1+A2+……+Am)+m(n-m)d
=Sm+m(n-m)d
=a+m(n-m)d=b
m(n-m)d=b-a
Sn=nA1+(1/2)n(n-1)d
(Sn)/n=A1+(n-1)d/2
(Sn)-b=S(n-m)=(n-m)A1+(1/2)(n-m)(n-m-1)d
(Sn-b)/(n-m)=A1+(n-m-1)d/2
两式相减
(Sn)/n-(Sn-b)/(n-m)=(n-1)d/2-(n-m-1)d/2=md/2
(Sn)×(n-m)/n-Sn+b=m(n-m)d/2
m(n-m)d=b-a
-(m/n)Sn=(b-a)/2-b=-(a+b)/2
Sn=n(a+b)/(2m)