已知Sn是首项为a的等比数列{an}的前n项和,S3,S9,S6成等差数列①求证:a2,a8,a5成等差数列;②若Tn=a1+2a4+3a7+……+na3n-2,求Tn
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![已知Sn是首项为a的等比数列{an}的前n项和,S3,S9,S6成等差数列①求证:a2,a8,a5成等差数列;②若Tn=a1+2a4+3a7+……+na3n-2,求Tn](/uploads/image/z/11680186-58-6.jpg?t=%E5%B7%B2%E7%9F%A5Sn%E6%98%AF%E9%A6%96%E9%A1%B9%E4%B8%BAa%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2CS3%2CS9%2CS6%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E2%91%A0%E6%B1%82%E8%AF%81%EF%BC%9Aa2%2Ca8%2Ca5%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%9B%E2%91%A1%E8%8B%A5Tn%3Da1%2B2a4%2B3a7%2B%E2%80%A6%E2%80%A6%2Bna3n-2%2C%E6%B1%82Tn)
已知Sn是首项为a的等比数列{an}的前n项和,S3,S9,S6成等差数列①求证:a2,a8,a5成等差数列;②若Tn=a1+2a4+3a7+……+na3n-2,求Tn
已知Sn是首项为a的等比数列{an}的前n项和,S3,S9,S6成等差数列
①求证:a2,a8,a5成等差数列;②若Tn=a1+2a4+3a7+……+na3n-2,求Tn
已知Sn是首项为a的等比数列{an}的前n项和,S3,S9,S6成等差数列①求证:a2,a8,a5成等差数列;②若Tn=a1+2a4+3a7+……+na3n-2,求Tn
1)因为an=a*q;Sn=a*(1-q^n)/(1-q);S3=a*(1-q^3)/(1-q);S6=a*(1-q^6)/(1-q);S9=a*(1-q^9)/(1-q);
2*S9=S3+S6;约去公约数a/(1-q)得 2*(1-q^9)=(1-q^3)+(1-q^6)
2*(1-q^3)(1+q^3+q^6)=(1-q^3)+(1-q^3)(1+q^3) 所以 2*(1+q^3+q^6)=1+1+q^3
所以 q^3= -0.5 令其=t;
a8=a*q^7=a*(-0.5)^2*q=0.25a*q;a2=a*q;a5=a*q^4= -0.5a*q;
所以 2*a8=a2+a5,即证a2,a8,a5成等差数列.
2)Tn=a1+2a4+3a7+……+na3n-2=a+2a*t+3*a*t^2+.+n*a*t^(n-1)
=[1+2t+3*t^2+.+n*t^(n-1)]*a ①
t*Tn=[ t+ 2t^2 +.+(n-1)*t^(n-1)+n*t^n]*a ②
①-②得 (1-t)Tn=[1+t+t^2+.+t^(n-1)-n*t^n]*a
因为t=-0.5 所以 Tn=(2/3)a*[2/3-(2/3+n)(-0.5)^n].