已知数列﹛an﹜的前n项和为Sn=3^n,数列﹛bn﹜满足b1=-1,bn+1=bn+(2n-1).已经得出n=1时,a1=3,n≥2时,an=2x3^(n-1);bn=n^2-2n;若cn=an*bn/n,求﹛cn﹜的前n项和Tn
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![已知数列﹛an﹜的前n项和为Sn=3^n,数列﹛bn﹜满足b1=-1,bn+1=bn+(2n-1).已经得出n=1时,a1=3,n≥2时,an=2x3^(n-1);bn=n^2-2n;若cn=an*bn/n,求﹛cn﹜的前n项和Tn](/uploads/image/z/4531910-14-0.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%B9%9Ban%EF%B9%9C%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%3D3%5En%2C%E6%95%B0%E5%88%97%EF%B9%9Bbn%EF%B9%9C%E6%BB%A1%E8%B6%B3b1%3D-1%2Cbn%2B1%3Dbn%2B%282n-1%29.%E5%B7%B2%E7%BB%8F%E5%BE%97%E5%87%BAn%3D1%E6%97%B6%2Ca1%3D3%2Cn%E2%89%A52%E6%97%B6%2Can%3D2x3%5E%28n-1%29%3Bbn%3Dn%5E2-2n%3B%E8%8B%A5cn%3Dan%2Abn%2Fn%2C%E6%B1%82%EF%B9%9Bcn%EF%B9%9C%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn)
已知数列﹛an﹜的前n项和为Sn=3^n,数列﹛bn﹜满足b1=-1,bn+1=bn+(2n-1).已经得出n=1时,a1=3,n≥2时,an=2x3^(n-1);bn=n^2-2n;若cn=an*bn/n,求﹛cn﹜的前n项和Tn
已知数列﹛an﹜的前n项和为Sn=3^n,数列﹛bn﹜满足b1=-1,bn+1=bn+(2n-1).已经得出n=1时,a1=3,n≥2时,
an=2x3^(n-1);bn=n^2-2n;若cn=an*bn/n,求﹛cn﹜的前n项和Tn
已知数列﹛an﹜的前n项和为Sn=3^n,数列﹛bn﹜满足b1=-1,bn+1=bn+(2n-1).已经得出n=1时,a1=3,n≥2时,an=2x3^(n-1);bn=n^2-2n;若cn=an*bn/n,求﹛cn﹜的前n项和Tn
由题知:b(n+1)=bn+(2n-1)
则:
b2=b1+(2*1-1)
b3=b2+(2*2-1)
b4=b3+(2*3-1)
.
bn=b(n-1)+[2*(n-1)-1]
累加法,得:bn=b1+(2*1-1)+(2*2-1)+(2*3-1)+.+[2*(n-1)-1]
=b1+2*[1+2+3+.+(n-1)]-(n-1)
=-1+n*(n-1)-(n-1)
=n²-2n
即:bn=n²-2n
又因为:当n>1时,an=2*[3^(n-1)]
且:cn=an*bn/n
则:当n=1时,cn=c1=a1*b1/1=-3
当n>1时,cn=(2n-4)*[3^(n-1)]
所以:Tn=-3+(2*2-4)*(3^1)+(2*3-4)*(3^2)+(2*4-4)*(3^3)+.+(2n-4)*[3^(n-1)] ①
则:
3Tn=-9+(2*2-4)*(3^2)+(2*3-4)*(3^3)+(2*4-4)*(3^4)+.+[2(n-1)-4]*[3^(n-1)]+(2n-4)*(3^n) ②
①-②得:-2Tn=6+2*(3^2)+2*(3^3)+2*(3^4)+.+2*[3^(n-1)]-(2n-4)*(3^n)
=6+2*[3^2+3^3+3^4+.+3^(n-1)]-(2n-4)*(3^n)
=6+2*{9*[3^(n-2)]/2}-(2n-4)*(3^n)
=6+9*[3^(n-2)]-(2n-4)*(3^n)
=6+(3^n)-9-(2n-4)*(3^n)
=(5-2n)*(3^n)-3
即:-2Tn=(5-2n)*(3^n)-3
所以:Tn=[(2n-5)*(3^n)+3]/2
bn用累加法求,
cn用错位相消法求解杰克