已知数列{cn}满足cn=3/bnxb(n+1),bn=3n-2.求数列{cn}的前n项和Tn
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已知数列{cn}满足cn=3/bnxb(n+1),bn=3n-2.求数列{cn}的前n项和Tn
cn=3/[bnb(n+1)]
=3/[(3n-2)(3(n+1)-2)]
=3/[(3n-2)(3n+1)]
=3×(1/3)×[1/(3n-2)-1/(3n+1)]
=1/(3n-2)-1/[3(n+1)-2]
Tn=c1+c2+...+cn
=1/(3×1-2)-1/(3×2-2)+1/(3×2-2)-1/(3×3-2)+...+1/(3n-2)-1/[3(n+1)-2]
=1/(3×1-2) -1/[3(n+1)-2]
=1 -1/(3n+1)
再问: 3×(1/3)×[1/(3n-2)-1/(3n+1)]解释一下
再答: 1/(3n-2) -1/[3(n+1)-2] =1/(3n-2)-1/(3n+1) =[(3n+1)-(3n-2)]/[(3n-2)(3n+1)] =3/[(3n-2)(3n+1)] 1/[(3n-2)(3n+1)]=(1/3)[1/(3n-2) - 1/[3(n+1)-2] ] 明白了吧。这是最基本的变形,会经常用到。 一般的:形如:1/[(an+b)(an+b+k)]=(1/k)[1/(an+b) - 1/(an +b+k)]
=3/[(3n-2)(3(n+1)-2)]
=3/[(3n-2)(3n+1)]
=3×(1/3)×[1/(3n-2)-1/(3n+1)]
=1/(3n-2)-1/[3(n+1)-2]
Tn=c1+c2+...+cn
=1/(3×1-2)-1/(3×2-2)+1/(3×2-2)-1/(3×3-2)+...+1/(3n-2)-1/[3(n+1)-2]
=1/(3×1-2) -1/[3(n+1)-2]
=1 -1/(3n+1)
再问: 3×(1/3)×[1/(3n-2)-1/(3n+1)]解释一下
再答: 1/(3n-2) -1/[3(n+1)-2] =1/(3n-2)-1/(3n+1) =[(3n+1)-(3n-2)]/[(3n-2)(3n+1)] =3/[(3n-2)(3n+1)] 1/[(3n-2)(3n+1)]=(1/3)[1/(3n-2) - 1/[3(n+1)-2] ] 明白了吧。这是最基本的变形,会经常用到。 一般的:形如:1/[(an+b)(an+b+k)]=(1/k)[1/(an+b) - 1/(an +b+k)]
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