
数学第三题要过程
2个回答
展开全部
)由anbn+1-an+1bn+2bn+1bn=0,cn=
an
bn
,可得数列{cn}是以1为首项,2为公差的等差数列,即可求数列{cn}的通项公式;
(2)用错位相减法来求和.
(1)∵anbn+1−an+1bn+2bn+1bn=0,cn=anbn,
∴cn−cn+1+2=0,
∴cn+1−cn=2,
∵首项是1的两个数列{an},{bn},
∴数列{cn}是以1为首项,2为公差的等差数列,
∴cn=2n−1;
(2)∵bn=3n−1,cn=anbn,
∴an=(2n−1)⋅31−n,
∴Sn=1×30+3×3−1+…+(2n−1)×31−n,
∴3Sn=1×3+3×30+…+(2n−1)×32−n,
∴−2Sn=−1×3−2⋅(30+3−1+…+32−n)−(2n−1)⋅31−n,
∴Sn=(n−1)31−n+3.
an
bn
,可得数列{cn}是以1为首项,2为公差的等差数列,即可求数列{cn}的通项公式;
(2)用错位相减法来求和.
(1)∵anbn+1−an+1bn+2bn+1bn=0,cn=anbn,
∴cn−cn+1+2=0,
∴cn+1−cn=2,
∵首项是1的两个数列{an},{bn},
∴数列{cn}是以1为首项,2为公差的等差数列,
∴cn=2n−1;
(2)∵bn=3n−1,cn=anbn,
∴an=(2n−1)⋅31−n,
∴Sn=1×30+3×3−1+…+(2n−1)×31−n,
∴3Sn=1×3+3×30+…+(2n−1)×32−n,
∴−2Sn=−1×3−2⋅(30+3−1+…+32−n)−(2n−1)⋅31−n,
∴Sn=(n−1)31−n+3.
展开全部
(1)
a1=b1=1
an.b(n+1)-a(n+1).bn+2b(n+1).bn=0
an.b(n+1)+2b(n+1).bn=a(n+1).bn
an+2bn=a(n+1)/b(n+1).bn
an/bn + 2 =a(n+1)/b(n+1)
a(n+1)/b(n+1)- an/bn = 2
=>{an/bn} 是等差数列, d=2
cn - c1 = 2(n-1)
cn -1 =2(n-1)
cn = 2n-1
(2)
bn=3^(n-1)
let
S = 1.3^0 +2.3^1+....+n.3^(n-1) (1)
3S = 1.3^1 +2.3^2+....+n.3^n (2)
(2)-(1)
2S = n.3^n -(3^0+3^1+...+3^(n-1))
=n.3^n - (1/2)(3^n -1)
cn = 2n-1
an = (2n-1) .3^(n-1)
Sn = a1+a2+...+an
= 2S - (1/2)(3^n -1)
=n.3^n - (1/2)(3^n -1) - (1/2)(3^n -1)
=n.3^n -(3^n -1)
=1+ (n-1).3^n
a1=b1=1
an.b(n+1)-a(n+1).bn+2b(n+1).bn=0
an.b(n+1)+2b(n+1).bn=a(n+1).bn
an+2bn=a(n+1)/b(n+1).bn
an/bn + 2 =a(n+1)/b(n+1)
a(n+1)/b(n+1)- an/bn = 2
=>{an/bn} 是等差数列, d=2
cn - c1 = 2(n-1)
cn -1 =2(n-1)
cn = 2n-1
(2)
bn=3^(n-1)
let
S = 1.3^0 +2.3^1+....+n.3^(n-1) (1)
3S = 1.3^1 +2.3^2+....+n.3^n (2)
(2)-(1)
2S = n.3^n -(3^0+3^1+...+3^(n-1))
=n.3^n - (1/2)(3^n -1)
cn = 2n-1
an = (2n-1) .3^(n-1)
Sn = a1+a2+...+an
= 2S - (1/2)(3^n -1)
=n.3^n - (1/2)(3^n -1) - (1/2)(3^n -1)
=n.3^n -(3^n -1)
=1+ (n-1).3^n
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