设M=1.02^1+1.02^2+1.02^3+……+1.02^64①
则1.02M=1.02^2+1.02^3+1.02^4+……+1.02^65②
②-①得:(1.02-1)M= 1.02^65-1.02^1
即M=( 1.02^65-1.02^1)/(1.02-1)
所以F=1400*(1.02^65-1.02^1)/(1.02-1)=70000(1.02^65-1.02)
设M=1.02^1+1.02^2+1.02^3+……+1.02^64①
则1.02M=1.02^2+1.02^3+1.02^4+……+1.02^65②
②-①得:(1.02-1)M= 1.02^65-1.02^1
即M=( 1.02^65-1.02^1)/(1.02-1)
所以F=1400*(1.02^65-1.02^1)/(1.02-1)=70000(1.02^65-1.02)