如何用数学归纳法证明这题目？

证明：
1.当n=1时，
左边=1*q^0=1
右边=[1-(1+1)q+q^2]/[(1-q)^2]=(1-q)^2/(1-q)^2=1
2.假设n=k时等式成立，即有：
1+2q+3q^2+…+kq^(k-1)=[1-(k+1)q^k+kq^(k+1)]/[(1-q)^2]
则n=k+1时，
左边=1+2q+3q^2+…+kq^(k-1)+(k+1)q^k
=[1-(k+1)q^k+kq^(k+1)]/[(1-q)^2]+(k+1)q^k
=[1-(k+1)q^k+kq^(k+1)+(k+1)q^k*(1-q)^2]/[(1-q)^2]
=[1-(k+1)q^k+kq^(k+1)+(k+1)q^k*(q^2-2q+1)]/[(1-q)^2]
=[1-(k+1)q^k+kq^(k+1)+(k+1)q^(k+2)-2(k+1)q^(k+1)+(k+1)q^k]/[(1-q)^2]
=[1-(k+2)q^(k+1)+(k+1)q^(k+2)]/[(1-q)^2]
故n=k+1时也成立
综上对任意n，均有1+2q+3q^2+…+nq^(n-1)=[1-(n+1)q^n+nq^(n+1)]/[(1-q)^2]