Question

求和：Sn=x+2x^2+3x^3+……+nx^x

Answer 1

解析：两边同乘以xxSn=x^2+2x^3+…nx^(n+1).①Sn=x+2x^2+3x^3+……+nx^x...②②-①则(1-x)*Sn=x+x^2+x^3+…+x^n-nx^(n+1)Sn=(x+x²+x³+...+x^n)/(1-x)=[ x(1-x^n)/(1-x)]/(1-x)=x(1-x^n)/(1-x)²...