an列方程组计算得a1=3,d=-1
an=4-n
bn=nq^(n-1)
sn=1+2q+3q^2+4q^3+.+nq^(n-1)
kn=n+(n-1)q+(n-2)q^2+(n-3)q^3+.+2q^(n-2)+q^(n-1)
=(1-q^n)/(1-q)+(1-q^(n-1))/(1-q)+.+(1-q^2)/(1-q)+(1-q)/(1-q)
=(n-(q+q^2+.+q^n))/(1-q)
=(n(1-q)-q+q^(n+1))/(1-q)^2
sn+kn=(n+1)(1-q^n)/(1-q)
sn=(n+1)(1-q^n)/(1-q)-kn=(1-nq^n-q^n+nq^(n+1))/(1-q)^2