已知an=1^2+2^2+3^2+.....+n^2求数列3/a1,5/a2,7/a3,......2n+1/an的前n项和
热心网友
已知an=1^2+2^2+3^2+......+n^2=n(n+1)(2n+1)/6---(2n+1)/an=6/[n(n+1)]---3/a1+5/a2+7/a3+......+(2n+1)/an=6/(1*2)+6/(2*3)+6/(3*4)+......+6/[n(n+1)]=6{(2-1)/(1*2)+(3-2)/(2*3)+(4-3)/(3*4)+......+[(n+1)-n]//[n(n+1)]}=6{(1-1/2)+(1/2-1/3)-(1/3-1/4)+......+[1/n-1/(n+1)]}=6[1-1/(n+1)]=6n/(n+1)
热心网友
an=n(n+1)(2n+1)/6(2n+1)/an=6/n(n+1)由裂项相消得3/a1,5/a2,7/a3,......（2n+1）/an=6[1-1/2+1/2-1/3+1/3-1/4+…+1/n - 1/（n+1）]=6-6/（n+1）