Question

高一数学，数列

Answer 1

1),a1=1 ,a2=3 ,a3=5
所以an=2n-1
Sn=n[1+(2n-1)]/2=n^2
Sn=n^2 ,S(n+1)=(n+1)^2
n[S(n+1)]^2-(n+1)(Sn)^2=n(n+1)^4-(n+1)*n^4
=n(n+1)[(n+1)^3-n^3]
=n(n+1)[(n+1)-n][(n+1)^2+n(n+1)+n^2]
=n(n+1)(3n^2+3n+1)
=(n+1)(3n^3+3n^2+n)
所以n[S(n+1)]^2-(n+1)(Sn)^2=(n+1)(3n^3+3n^2+n)=(n+1)(3n^3+An^2+Bn)
A=3 ,B=1
an=an=2n-1
2)
(n+1)an=b1/2+b2/2^2+...+bn/2^n
na(n-1)=b1/2+b2/2^2+...+b(n-1)/2^(n-1)
两式相减得
(n+1)an-na(n-1)=bn/2^n
(n+1)*(2n-1)-n*(2n-3)=bn/2^n
bn=2^n[(n+1)*(2n-1)-n*(2n-3)]=(4n-1)*2^n
bn=(4n-1)*2^n
Tn=3*2+7*2^2+11*2^3+...+(4n-1)*2^n
2Tn=3*2^2+7*2^3+...+(4n-5)*2^n+(4n-1)*2^(n+1)
两式相减得
-Tn=3*2+4[2^2+2^3+...+2^n]-(4n-1)*2^(n+1)
-Tn=3*2+4[4(1-2^(n-1)]/(1-2)-(4n-1)*2^(n+1)
-Tn=3*2+4[-4(1-2^(n-1)]-(4n-1)*2^(n+1)
-Tn=-10+8*2^n-2(4n-1)*2^n
Tn=10+(8n-10)*2^n

追问

如何知道它是等差

Answer 2

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