求和:Sn=1-3x+5x^2-7x^3+.+(2n+1)(-x)^n(n属于N*)
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求和:Sn=1-3x+5x^2-7x^3+.+(2n+1)(-x)^n(n属于N*)
Sn 是等差数列an = 2n-1,等比数列 bn = (-x)^(n-1)
前n+1的和
Sn -(-x) Sn = (a1b1 + a2b2+...an+1 bn+1) - (a1b2 + a2b3+ ...+ an+1 bn+2)
= a1b1 + [ (a2-a1)b2 + (a3-a2)b3 +...+ (an+1 - an)bn+1] - an+1 bn+2
= 1 + 2*[ b2+b3 +...+bn+1] - (2n+1)bn+2
= 1 + 2*[ b2+b3 +...+bn+1] - (2n+1)(-x)^(n+1)
= 1 + 2* [(b2-bn+2)/(1-(-x))] - (2n+1)(-x)^(n+1)
= 1 + 2*[ (-x - (-x)^(n+1))/(1+x) ] - (2n+1)(-x)^(n+1)
= 1 + 2(-x)/(1+x) - 2(-x)^(n+1)/(1+x) - (2n+1)(-x)^(n+1)
Sn = (Sn-(-x)Sn)/(1+x)
= 1/(1+x) -2x/(1+x)^2 - 2(-x)^(n+1)/(1+x)^2 - (2n+1)(-x)^(n+1)/(1+x)
前n+1的和
Sn -(-x) Sn = (a1b1 + a2b2+...an+1 bn+1) - (a1b2 + a2b3+ ...+ an+1 bn+2)
= a1b1 + [ (a2-a1)b2 + (a3-a2)b3 +...+ (an+1 - an)bn+1] - an+1 bn+2
= 1 + 2*[ b2+b3 +...+bn+1] - (2n+1)bn+2
= 1 + 2*[ b2+b3 +...+bn+1] - (2n+1)(-x)^(n+1)
= 1 + 2* [(b2-bn+2)/(1-(-x))] - (2n+1)(-x)^(n+1)
= 1 + 2*[ (-x - (-x)^(n+1))/(1+x) ] - (2n+1)(-x)^(n+1)
= 1 + 2(-x)/(1+x) - 2(-x)^(n+1)/(1+x) - (2n+1)(-x)^(n+1)
Sn = (Sn-(-x)Sn)/(1+x)
= 1/(1+x) -2x/(1+x)^2 - 2(-x)^(n+1)/(1+x)^2 - (2n+1)(-x)^(n+1)/(1+x)
求和:Sn=1-3x+5x^2-7x^3+.+(2n+1)(-x)^n(n属于N*)
\求和Sn=1+2x+3x^2+```+(n-1)x^(n-2)+n*x^(n-1)
求和:Sn=1+3x+5x+7x+...+(2n-1)x^(n-1)谢谢了,
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