Question

{an}是等差数列前n项和Sn已知Sm=a Sn-Sn-m=b 求Sn

Answer 1

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)