1^2*q^1+2^2*q^2+3^2*q^3+4^2*q^4+...+n^2*q^n=?
令S=1^2*q^1+2^2*q^2+3^2*q^3+4^2*q^4+...+n^2*q^n…………(1);
两边同乘以q:
qS=1^2*q^2+2^2*q^3+3^2*q^4+4^2*q^5+...+n^2*q^(n+1)…………(2);
(1)-(2)得:
S-qS=(1-q)S=q+3*q^2+5*q^3+7*q^4+9*q^5+...+(2n-1)*q^n-n^2*q^(n+1)
两边同乘以q:
q(1-q)S=q^2+3*q^3+5*q^4+7*q^5+9*q^6+...+(2n-1)*q^(n+1)-n^2*q^(n+2)…………(3);
(2)-(3)得:
S(1-q)^2=2(q+q^2+q^3+q^4+q^5+...+q^n)-q-(n^2+2n-1)*q^(n+1)-n^2*q^(n+2)
=2(q(1-q^n)/(1-q)-q-(n^2+2n-1)*q^(n+1)-n^2*q^(n+2)
所以得:
S=[2(q(1-q^n)/(1-q)-q-(n^2+2n-1)*q^(n+1)-n^2*q^(n+2)]/(1-q)^2
后面就不化简了,我计算能力差