Question

数列的前n项和sn=n^2+2n 求数列的通项公式an

Answer 1

sn=n^2+2n

s(n-1)=(n-1)^2+2(n-1)=n^2-1

所以an=sn-s(n-1)=(n^2+2n)-(n^2-1)=2n+1

a1=s1=1+2=3

符合an=2n+1

所以an=2n+1

a1=3,a2=5,a3=7,……

所以Tn=1/3*5+1/5*7+……+1/(2n+1)(2n-1)

=[2/3*5+2/5*7+……+2/(2n+1)(2n-1)]/2

={(1/3-1/5)+(1/5-17)+……+[1/(2n-1)-1/(2n+1)]}/2

=[1/3-1/(2n+1)]/2

=(n-1)/(2n+1)