已知等差数列{an}满足∶a3=7,a5+a7=26,{an}的前几项和为sn令bn=2的n次方乘an,求数列前n项和Tn
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![已知等差数列{an}满足∶a3=7,a5+a7=26,{an}的前几项和为sn令bn=2的n次方乘an,求数列前n项和Tn](/uploads/image/z/2427545-65-5.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3%E2%88%B6a3%3D7%2Ca5%2Ba7%3D26%2C%7Ban%7D%E7%9A%84%E5%89%8D%E5%87%A0%E9%A1%B9%E5%92%8C%E4%B8%BAsn%E4%BB%A4bn%3D2%E7%9A%84n%E6%AC%A1%E6%96%B9%E4%B9%98an%2C%E6%B1%82%E6%95%B0%E5%88%97%E5%89%8Dn%E9%A1%B9%E5%92%8CTn)
已知等差数列{an}满足∶a3=7,a5+a7=26,{an}的前几项和为sn令bn=2的n次方乘an,求数列前n项和Tn
已知等差数列{an}满足∶a3=7,a5+a7=26,{an}的前几项和为sn令bn=2的n次方乘an,求数列前n项和Tn
已知等差数列{an}满足∶a3=7,a5+a7=26,{an}的前几项和为sn令bn=2的n次方乘an,求数列前n项和Tn
∵a5 = a3 + 2d
a7 = a3 + 4d
∴a5+a7= 2a3 + 6d = 26
又∵a3 = 7
∴d = 2
∴a1 = a3 - 2d = 3
∴an = a1 + (n-1)d = 3 + 2(n-1) = 2n + 1
∴bn = (2n+1)×2^n
Tn = 3×2 + 5×2^2 + 7×2^3 + …… + (2n+1)×2^n
2Tn = 3×2^2 + 5×2^3 + …… + (2n-1)×2^n + (2n+1)×2^(n+1)
两式相减,得:
Tn = (2n+1)×2^(n+1) - 3×2 - [2^3 + 2^4 + …… + 2^(n+1)]
= (2n+1)×2^(n+1) - 6 - 8[1-2^(n-1)]/(1-2)
= (2n+1)×2^(n+1) - 6 - 8×2^(n-1) + 8
= (2n+1)×2^(n+1) - 6 - 2×2^(n+1) + 8
= (2n-1)×2^(n+1) + 2
a3=7 a5+a7=26
so,an=6n-11
bn=2^n*an=2^n(6n-11)
Tn=b1+b2+⋯⋯+bn然后用裂项求和就可以了详细一点的过程可以不....这里学的不是特别好,看过程我会更明白一些,麻烦了Tn=2(6-11)+2^2(6*2-11)+2^3(6*3-11)+⋯⋯+2^(n-1)(6(n-1)-...
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a3=7 a5+a7=26
so,an=6n-11
bn=2^n*an=2^n(6n-11)
Tn=b1+b2+⋯⋯+bn然后用裂项求和就可以了
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