兄弟,我來幫你
錯位相減法
bn=-n•2^n
b1=-1•2^1,b2=-2•2^2,b3=-3•2^3,...,bn-1=-(n-1)•2^(n-1),bn=-n•2^n
Sn=b1+b2+b3+...+bn-1+bn
Sn=-1•2^1-2•2^2-3•2^3-...-(n-1)•2^(n-1)-n•2^n.①
2Sn=-1•2^2-2•2^3-3•2^4-...-(n-1)•2^n-n•2^(n+1).②
②-①得
Sn=1•2^1+1•2^2+1•2^3+1•2^n-n•2(n+1)
Sn=2^1+2^2+2^3+...+2^n-n•2^(n+1)
Sn=2•(1-2^n)/(1-2)-n•2^(n+1)
Sn=(1-n)•2^(n+1)-2
錯位相減法
若數(shù)列{an}是等差數(shù)列,數(shù)列{bn}是等比數(shù)列,由這兩個數(shù)列的對應(yīng)項的乘積組成的新數(shù)列{anbn},當(dāng)求數(shù)列的前n項和時,常常采用將{anbn}各項乘以{bn}的公比q,并向后錯一項與原{anbn}的同次項對應(yīng)相減的方法.
例:求數(shù)列1,2x,3x^2,4x^3,...,nx^(n-1)前n項之和
Sn=1+2x+3x^2+4x^3+...+nx^(n-1)...①
xSn=x+2x^2+3x^3+4x^4+...+nx^n...②
①-②得
(1-x)Sn=1+x+x^2+x^3+...+x^(n-1)-nx^n
當(dāng)x≠1時
(1-x)Sn=(1-x^n)/(1-x)-nx^n
Sn=(1-x^n)/(1-x)^2-nx^n/(1-x)=[1-(1+n)x^n+nx^n]/(1-x)^2
當(dāng)x=1時
原式中Sn=1+2+3+...+n=n(1+n)/2
例:求數(shù)列1/2,3/4,5/8,...,(2n-1)/2^n的前n項之和
Sn=1/2+3/4+5/8+7/16+...+(2n-3)/2^(n-1)+(2n-1)/2^n...①
Sn/2=1/4+3/8+5/16+...+(2n-3)/2^n+(2n-1)/2^(n+1)...②
①-②得
Sn-Sn/2=1/2+(3/4-1/4)+(5/8-3/8)+(7/16-5/16)+...+[(2n-1)/2^n-(2n-3)/2^n]-(2n-1)/2^(n+1)
Sn/2=1/2+(2/4+2/8+2/16+...+2/2^n)-(2n-1)/2^(n+1)
Sn/2=1/2+[1/2+1/4+1/8+...+1/2^(n-1)]-(2n-1)/2^(n+1)
Sn/2=1/2+{(1/2)·[1-1/2^(n-1)]/(1-1/2)}-(2n-1)/2^(n+1)
Sn/2=1/2+1-1/2^(n-1)-(2n-1)/2^(n+1)
Sn/2=3/2-[4/2^(n+1)+(2n-1)/2^(n+1)]
Sn/2=3/2-(2n+3)/2^(n+1)
Sn=3-(2n+3)/2^n