已知为an满足a1=1,a2=2且an 1=2an 3an-1

an=1＋2＋3＋…＋n=[n(n＋1)]/2则：1/(an)=2/[n(n＋1)]=2[(1/n)－1/(n＋1)],所以：M=1/(a1)＋1/(a2)＋1/(a3)＋…＋1/(an)=2[1/1

a1+2a2+3a3+...+nan=n(n+2)(n+1)a1+2a2+3a3+...+(n-1)a(n-1)=(n-1)n(n+1)nan-(n-1)a(n-1)=3n(n+1)nan=(n-1)

设公差为d,公比为q,则b2=qb1=q(a1+1)=(a1+d+2),↔2q=3+d,b3=q²b1=q²(a1+1)=(a1+2d+3),↔q²

由an+2=3an+1-2an可得an+2-an+1=2(an+1-an)因为a2-a1=2,所以an+1-an不会等于0,则an+1-an是以2为公比的等比数列由上可得an+1-an=2^nan-a

a1+d=97a1+4d=85解得a1=101,d=-4101-4(k-1)>0,101-4(k+1-1)

1+2+3+.+n-1=(1+n-1)(n-1)/2等差数列求和哦~所以跟外面的2约了!

（1）证b1=a2-a1=1，当n≥2时，bn＝an+1−an＝an−1+an2−an＝−12(an−an−1)＝−12bn−1，所以{bn}是以1为首项，−12为公比的等比数列．（2）解由（1）知b

an/a(n-1)=1×2^(n-1)=2^(n-1)a(n-1)/a(n-2)=2^(n-2)…………a2/a1=2连乘an/a1=2×2²×...×2^(n-1)=2^[1+2+...+

令Sn为an前n项和,Sn=n-an,S(n-1)=n-1-a(n-1),两式相减,an=1-an+a（n-1）,2(an-1)=a(n-1)-1,所以an-1是公比为1/2的等比数列,a1-1=-1

∵a2*a4=4∴a3=2.q=1/2.an=2^(4-n)2^(9-3n)>1/9.9-3n>=-3n

1.a1+3a2+...+3^(n-1)an=n/3①a1+3a2+...+3^(n-2)a(n-1)=(n-1)/3②①-②3^(n-1)an=n/3-(n-1)/3=1/3an=(1/3)^n2.

a1(q+q^3)=4a1(1+q+q^2)=14两式相除：(q+q^3)/(1+q+q^2)=2/7求得qan+an+1+an+2=（a1+a2+a3）*q^(n-1)>1/9关键是求q说实在的,我

a1+a2+a3+...+an=n^2+2n可得：Sn=a1+a2+a3+...+an=n^2+2n当n=1时有：a1=S1=1+2=3当n≥2时有：an=Sn-S(n-1)=n^2+2n-(n-1)

（I）∵bn=an+1-an，∴an+2-2an+1+an=bn+1-bn=2n-6∴bn−bn−1＝2(n−1)−6，bn−1−bn−2＝2(n−2)−6，…，b2−b1＝2−6将这n-1个等式相加

a(n+1)/an=1×2ⁿ=2ⁿan/a(n-1)=2^(n-1)a(n-1)/a(n-2)=2^(n-2)…………a2/a1=2连乘an/a1=2×2²×...×

a(n+2)=3*a(n+1)-2*ana(n+2)-a(n+1)=2*(a(n+1)-an)a2-a1=3-1=2a(n+1)-an=2^na(n+2)-2a(n+1)=a(n+1)-2*ana2-

a2=a1+2a2=1+2a2得a2=-1an=a1+2a2+3a3+...+(n-2)a(n-2)＋(n-1)a(n-1)a(n-1)＝a1+2a2+3a3+...+(n-2)a(n-2)两式相减：

经化简得a1a2a3分别为a1=4a2=a1+3p+1=5+3p a3=a1+12p+2=6+12pa1,a2+6,a3成等差数列.的2a2+12=a1+a3即22+6p=10+12p解得p

an=n

a(3)=a(1+2)=1/[1+a(1)]=a(1),1=a(1)+[a(1)]^2,0=[a(1)]^2+a(1)-1,Delta=1+4=5.a(1)=[-1+5^(1/2)]/2,或a(1)=