求和1x2+3x2^2+5x2^3+…+(2n-1)x2n要详解答案,急用!
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求和1x2+3x2^2+5x2^3+…+(2n-1)x2n要详解答案,急用!求和1x2+3x2^2+5x2^3+…+(2n-1)x2n要详解答案,急用!求和1x2+3x2^2+5x2^3+…+(2n-
求和1x2+3x2^2+5x2^3+…+(2n-1)x2n要详解答案,急用!
求和1x2+3x2^2+5x2^3+…+(2n-1)x2n要详解答案,急用!
求和1x2+3x2^2+5x2^3+…+(2n-1)x2n要详解答案,急用!
S=1x2+3x2^2+5x2^3+…+(2n-1)x2^n
2S=1x2^2+3x2^3+5x2^4+…+(2n-3)x2^n+(2n-1)x2^(n+1)
2S-S=S=-2-2x2^2-2x2^3-.-2x2^n+(2n-1)x2^(n+1)
= -2-2^3-2^4-.-2^(n+1)+ (2n-1)x2^(n+1)
设Tn=1x2+3x2^2+5x2^3+…+(2n-1)x2n
2Tn=1x2^2+3x2^2+…+(2n-1)x2^n+1
Tn=2+2^2+2^3+…+2^n=2^n-2
用错位相减法。
Sn=1×2+3×2^2+5×2^3+...+(2n-1)×2^n
2Sn=1×2^2+3×2^3+...+(2n-3)×2^n+(2n-1)×2^n
Sn-2Sn=-Sn=-2+2×2+2×2^2+...+2×2^n-(2n-1)×2^n=-2+2(2^n-1)/(2-1)-(2n-1)×2^n
Sn=(2n-3)×2^n+4
错位相减很容易可以得到Sn=(2n+1)×2^n+4
将这个数列乘2,两个数列错位相减,即带相同2的次方项的两个数相减。。然后就好求了
An=(2n-1)x2^n=nx2^(n+1)-2^n,
则Sn=[nx2^2x(2^n-1)/(2-1)]-[2x(2^n-1)/2-1]=(2^n-1)(4n-1)
求和Sn=1x2+3x2^2+5x2^3+…+(2n-1)x2^n
求和1x2+3x2^2+5x2^3+…+(2n-1)x2n要详解答案,我采纳!急用!
求和1x2+3x2^2+5x2^3+…+(2n-1)x2n要详解答案,急用!
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