两道有点难度的数列题，高手指点一下
\u4e24\u9053\u6709\u5173\u6570\u5217\u7684\u9898\u76ee\u4e00.
\u89e3\uff1a
1
(ak2)^2=(ak1)*(ak31)
\u5373\uff1a(a7)^2=(a1)*(a31)
\u5373\uff1a(a1+6d)^2=(a1)*(a1+30d)
\u56e0\u4e3ad!=0,\u6240\u4ee5a1=2*d; (!=\u662f\u4e0d\u7b49\u4e8e)
\u6240\u4ee5an=(n+1)d
\u8bbe\u7b49\u6bd4\u6570\u5217\u7684\u516c\u6bd4\u4e3a\uff1aq
q=a2/a1=3/2
ak1=2*d
\u6240\u4ee5akn=(ak1)*(3/2)^(n-1)=2d*(3/2)^(n-1)
\u4e14akn=((kn)+1)*d
\u6240\u4ee52d*(3/2)^(n-1)=((kn)+1)*d
\u6240\u4ee5kn=2(3/2)^(n-1)-1
2
Tn=2*((3/2)^n-1)/(3/2-1)-n=4*((3/2)^n-1)-n
\u4e8c
1
Sn=-an-\uff081/2\uff09^\uff08n-1\uff09+2 (1)
S(n-1)=-a(n-1)-(1/2)^(n-2)+2 (2)
(1)-(2) \u5f97\uff1a
an=a(n-1)-an+(1/2)^(n-2)-\uff081/2\uff09^\uff08n-1)
\u5316\u7b80\u5f97\uff1a
2an=a(n-1)+(1/2)^(n-1)
\u6240\u4ee52^n*an=2^(n-1)*a(n-1)+1
\u6240\u4ee5bn=b(n-1)+1
b1=2*a1=2*S1=1
\u6240\u4ee5bn=n
2
an=n/(2^n)
cn=(n+1)/(2^n)
Tn= c1+c2+...cn
= 1/2+2/2^2+... +n/2^n (1)
Tn/2= 1/2^2+2/2^3+...+(n-1)/2^n+n/2^(n+1) (2)
(1)-(2) \u5f97\uff1a
Tn/2=1/2+1/2^2+...+1/2^n-n/2^(n+1)
=1/2*(1-(1/2)^n)/(1-1/2)-n/2^(n+1)
=1-(n+2)/2^(n+1)
Tn=2-(n+2)/2^n
n->\u65e0\u7a77\u5927\u65f6 Tn<\uff085n\uff09/\uff082n+1\uff09\u662f\u4e00\u5b9a\u7684
\u81f3\u4e8e\u4e2d\u95f4\u6709\u6ca1\u6709 Tn>\uff085n\uff09/\uff082n+1\uff09\u7684\u60c5\u51b5\uff0c
\u4f60\u81ea\u5df1\u7814\u7a76\u5427
\u4ec5\u505a\u53c2\u8003
\u7b2c\u4e00\u4e2a\uff1a
an=7*(10^0+10^1+10^2+\u2026\u2026+10^(n-1))=7*(10^n-1)/9
\u7b2c\u4e8c\u4e2a\uff0c\u51494\u9879\u4e0d\u597d\u770b\u51fa\u89c4\u5f8b\uff0c\u6216\u8005\u8bf4\u53ef\u80fd\u4f1a\u4ea7\u751f\u5f88\u591a\u79cd\u89e3
待定系数法：a(n+1)-1=-1/3(an-1)
得an-1=8(-1/3)^(n-1)
所以an=8(-1/3)^(n-1)+1
Sn=6(1-(-1/3)^(n)+n
|Sn-n-6|=|6(-1/3)^n|=|6*(1/3)^n|<1/125
n>=6
2.我感觉是累加，试验了下，果然是
利用an=An-A(n-1)；bn=Bn-B(n-1)代换
则Cn=AnBn-A(n-1)B(n-1)(n>=2)
C1=a1b1=A1B1
故C1+C2+C3+……+C10=A1B1+A2B2-A1B1+A3B3-A2B2+……+A10B10-A9B9
=A10B10
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