急求解答一道高一数学题设等差数列{an}的前n项和为Sn,且Sn=2分之1n*an+an-c(c是常数,n属于正整数),求c的值及{an}的通项公式
2019-05-23
急求解答一道高一数学题
设等差数列{an}的前n项和为Sn,且Sn=2分之1n*an+an-c(c是常数,n属于正整数),求c的值及{an}的通项公式
优质解答
因an=a1+(n-1)d
所Sn=2分之1n*an+an-c=n*[a1+(n-1)d]/2+a1+(n-1)d-c=(n+2)a1/2+(n+2)(n-1)d/2-c
因Sn=na1+n(n-1)d
所(n+2)a1/2+(n+2)(n-1)d/2-c=na1+n(n-1)d
整理得：c=(2-n)a1/2+(n-1)d
因an=a1+(n-1)d
所Sn=2分之1n*an+an-c=n*[a1+(n-1)d]/2+a1+(n-1)d-c=(n+2)a1/2+(n+2)(n-1)d/2-c
因Sn=na1+n(n-1)d
所(n+2)a1/2+(n+2)(n-1)d/2-c=na1+n(n-1)d
整理得：c=(2-n)a1/2+(n-1)d