求和:x+3x^2+5x^3+...+(2n-1)x^n
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求和:x+3x^2+5x^3+...+(2n-1)x^n
解析:
当x=0时,x+3x^2+5x^3+...+(2n-1)x^n=0;
当x=1时,x+3x^2+5x^3+...+(2n-1)x^n=1+3+5+...+(2n-1)=n(1+2n-1)/2=n²
当x≠0且x≠1时,
令S=x+3x^2+5x^3+...+(2n-1)x^n,那么:
xS=x²+3x³+5x^4 +7x^5 +...+(2n-3)x^n +(2n-1)x^(n+1)
所以:S-xS
=x+3x^2+5x^3+...+(2n-1)x^n -[x²+3x³+5x^4 +7x^5 +...+(2n-3)x^n +(2n-1)x^(n+1) ]
=x+2(x²+x³+x^4 +x^5 +...+x^n) -(2n-1)x^(n+1) (第二项括号内利用等比数列求和公式求和)
=x+2x²[1-x^(n-1)]/(1-x) -(2n-1)x^(n+1)
=[x-x²+2x²-2*x^(n+1) -(2n-1)x^(n+1) +(2n-1)x^(n+2)]/(1-x)
=[x+x²-(2n+1)x^(n+1) +(2n-1)x^(n+2)]/(1-x)
则S=[x+x²-(2n+1)x^(n+1) +(2n-1)x^(n+2)]/(1-x)²
等差数列和等比数列乘积构成的新数列求和 用错位相减法
设所求为Sn
则xSn=x^2+3x^3+5x^5+……+(2n-1)x^(n+1)
两式相减,则(1-x)Sn=x+2x^2+2x^3+……+2x^n-2x^(n+1)
然后等式右边是等比数列,对它求和,然后就会做了吧
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