【急.已知数列{(2n-1)·2^n},求其前N项和Sn利用错项相减求出Sn-2Sn=2+2^3+2^4+...+2^(n+1)- (2n-1)*2^n+1所以Sn =6-2^(n+2)*(2-n)为什么再代入值验算时都不对,
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![【急.已知数列{(2n-1)·2^n},求其前N项和Sn利用错项相减求出Sn-2Sn=2+2^3+2^4+...+2^(n+1)- (2n-1)*2^n+1所以Sn =6-2^(n+2)*(2-n)为什么再代入值验算时都不对,](/uploads/image/z/650381-5-1.jpg?t=%E3%80%90%E6%80%A5.%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7B%282n-1%29%C2%B72%5En%7D%2C%E6%B1%82%E5%85%B6%E5%89%8DN%E9%A1%B9%E5%92%8CSn%E5%88%A9%E7%94%A8%E9%94%99%E9%A1%B9%E7%9B%B8%E5%87%8F%E6%B1%82%E5%87%BASn-2Sn%3D2%2B2%5E3%2B2%5E4%2B...%2B2%5E%28n%2B1%29-+%282n-1%29%2A2%5En%2B1%E6%89%80%E4%BB%A5Sn+%3D6-2%5E%28n%2B2%29%2A%282-n%29%E4%B8%BA%E4%BB%80%E4%B9%88%E5%86%8D%E4%BB%A3%E5%85%A5%E5%80%BC%E9%AA%8C%E7%AE%97%E6%97%B6%E9%83%BD%E4%B8%8D%E5%AF%B9%2C)
【急.已知数列{(2n-1)·2^n},求其前N项和Sn利用错项相减求出Sn-2Sn=2+2^3+2^4+...+2^(n+1)- (2n-1)*2^n+1所以Sn =6-2^(n+2)*(2-n)为什么再代入值验算时都不对,
【急.已知数列{(2n-1)·2^n},求其前N项和Sn
利用错项相减求出
Sn-2Sn=2+2^3+2^4+...+2^(n+1)- (2n-1)*2^n+1
所以Sn =6-2^(n+2)*(2-n)
为什么再代入值验算时都不对,
【急.已知数列{(2n-1)·2^n},求其前N项和Sn利用错项相减求出Sn-2Sn=2+2^3+2^4+...+2^(n+1)- (2n-1)*2^n+1所以Sn =6-2^(n+2)*(2-n)为什么再代入值验算时都不对,
这个式子是对的Sn-2Sn=2+2^3+2^4+...+2^(n+1)- (2n-1)*2^n+1
然后是这样计算的:
-Sn=2+【2^3+2^4+...+2^(n+1)】- (2n-1)*2^(n+1) 中括号内有n-1项的等比数列求和
= 2+ 8【2^(n-1)-1】- (2n-1)*2^(n+1)
= -6-(2n-3)2^(n+1)
所以 Sn = 6 + (2n-3)2^(n+1)
an = n2^(n+1) - 2^n
令bn = n2^(n+1),cn = 2^n
bn的前n项为Bn,cn的前n项为Cn,an的前n项为Sn,
Bn = 1*2^2+2*2^3+3*2^4+......+n2^(n+1)
2Bn= 1*2^3+2*2^4+.......+(n-1)2^(n+1)+n2^(n+2)
相减:
-B...
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an = n2^(n+1) - 2^n
令bn = n2^(n+1),cn = 2^n
bn的前n项为Bn,cn的前n项为Cn,an的前n项为Sn,
Bn = 1*2^2+2*2^3+3*2^4+......+n2^(n+1)
2Bn= 1*2^3+2*2^4+.......+(n-1)2^(n+1)+n2^(n+2)
相减:
-Bn = 2^2+2^3+.......+2^(n+1)-n2^(n+2)
=4(2^n - 1)-n2^(n+2)
Bn = (n-1)2^(n+2) + 4
Cn = 2^(n+1)-2
Sn = Bn + Cn
=(n-1)2^(n+2) +2^(n+1) + 2
收起
-Sn=2+【2^3+2^4+...+2^(n+1)】- (2n-1)*2^(n+1) 中括号内有n-1项的等比数列求和
= 2+ 8【2^(n-1)-1】- (2n-1)*2^(n+1)
= -6-(2n-3)2^(n+1)
Sn = 6 + (2n-3)2^(n+1)
有正解了,飘过