二道数学题~~

设X=1+1/3+1/3^2+…+1/3^10
3X-X=(3+1+1/3+1/3^2+…+1/3^9)-(1+1/3+1/3^2+…+1/3^10)
2X=3-1/3^10
X=(3-1/3^10)/2=(3-1/59049)/2
=1又29524/59049

设X=1+2/3+3/3^2+4/3^3+…+11/3^10
(3^10)X=1/3^10+(2×3)/(3^10)+(3×3^2)/(3^10)+…+11×3^10/3^10
=[1+2×3+3×3^2+…+11×3^10]/(3^10)=
[1+6+27+108+405+1458+5103+17496+59049
+196830+649539]/(3^10)=930022/59049
=15又44287/59049

属于等比数列求和问题
设a=1+1/3+1/3^2+…+1/3^10
a=(a- 1/3a)*3/2
= [(1+1/3+1/3^2+…+1/3^10)-(1/3+1/3^2+…
+1/3^10+1/3^11)]*3/2
=(3-1/3^10)*1/2

设a=1+2/3+3/3^2+4/3^3+…+11/3^10
a=(a- 1/3a)*3/2
=[(1+2/3+3/3^2+4/3^3+…+11/3^10)-
(0+1/3+2/3^2+3/3^3+…+10/3^10+11/3^11)]*3/2
=[1-11/3^11+(1/3+1/3^2+1/3^3+…+1/3^10)]*3/2
=[1-11/3^11+((3-1/3^10)*1/2]*3/2
=15/4-1/2*11/3^10-1/2*1/3^7