数列{an}为等比数列,a1=1,a5=256,Sn为等差数列{bn}的前n项和,b1=2,5S5=2S8.1,求通项an,bn2,求{an+bn}的前n项和Tn
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数列{an}为等比数列,a1=1,a5=256,Sn为等差数列{bn}的前n项和,b1=2,5S5=2S8.1,求通项an,bn2,求{an+bn}的前n项和Tn
数列{an}为等比数列,a1=1,a5=256,Sn为等差数列{bn}的前n项和,b1=2,5S5=2S8.
1,求通项an,bn
2,求{an+bn}的前n项和Tn
数列{an}为等比数列,a1=1,a5=256,Sn为等差数列{bn}的前n项和,b1=2,5S5=2S8.1,求通项an,bn2,求{an+bn}的前n项和Tn
1) {an}为等比数列,a1=1,a5=256,
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...
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1) {an}为等比数列,a1=1,a5=256,
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
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