前面是等差数列,后面是等比数列.求和用"错位相减".
Sn=a1*b+(a1+d)bq+(a1+2d)bq^2+...(a1+(n-1)d)bq^(n-1)
qSn=a1bq+(a1+d)bq^2+(a1+3d)bq^3+...+(a1+(n-1)d)bq^n
二式相减得:
(1-q)Sn=a1b+db[q+q^2+...+q^(n-1)]-[a1+(n-1)d]bq^n
当q=1时,Sn=b[na1+1/2n(n-1)d]
当q不=1时:
Sn={a1b+db*q(1-q^(n-1))/(1-q)-[a1+(n-1)d]bq^n}/(1-q)