本学渣被这题难倒了,急需过程,学霸们帮帮忙,拜托了!急!!!
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发布时间：2024-10-21 22:18
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热心网友
时间：2024-10-22 00:42
令an=a1q^(n-1)=q^(n-1),
=> q=(a5/a1)^(1/4)=±4
∵an是为正数的等比数列
∴q=4，an=4^(n-1);
令bn=b1+(n-1)d=2+(n-1)d,
=> Sn=b1n+n(n-1)d/2=2n+n(n-1)d/2,
=> S5=10+10d, S8=16+28d,
5S5=2S8,
=> d=3,
=> bn=2+3(n-1)=3n-1
2)Tn=2×4^0+5×4^1+..+(3n-1)×4^(n-1)
∴4Tn=2×4^1+5×4^2+..+(3n-4)×4^(n-1)+(3n-1)×4^n
两式相减得：-3Tn=2×4^0+3×(4^1+4^2+...+4^(n-1))-(3n-1)×4^n
=2+3×4×[1-4^(n-1)]/(1-4)-(3n-1)×4^n
=2-4+4^n-(3n-1)×4^n
=-2+(2-3n)×4^n
∴Tn=2/3+(3n-2)/3×4^n
