它没有按预期工作。我该如何修复此代码？

Question

我希望代码每六个月增加一次我的semi_annual_income，方法是使semi_annual income 每六个月增加一个特定的百分比。所以假设根据我的数学1000(1.2)^i/6，这个等式将每 6 个月增加我的收入0.2，其中i 可以被 6 整除。

当我使用表达式1000(1.2)^i/6 时，我尝试了这两个表达式，它会给我一个非常大的数字。而表达式1000(1 +0.2) 给了我1000(1 + 0.2) 应该给我的确切答案。

number_of_months = 0
total_cost = 100000
semi_annual_income = 1000
starting_salary = 10
semi_annual_rise = 0.2
while semi_annual_income < total_cost:
    for i in range(100):
        if float(i)%6 == 0:
            power_number = float(i)/6
            # i am using this to make it increase the semi_annual income just only every six months
            semi_annual_income = semi_annual_income *(float(1) + float(semi_annual_rise))
            print(power_number)
            print(semi_annual_income)
            #semi_annual_income = (semi_annual_income *(float(1) +  float(semi_annual_rise))** float(power_number))
            #The above written code is giving me a very huge number i  want it to give me the answer as 1000(1 + 0.2)^i/6
    break

我从代码中得到了我想要的答案，但我不明白为什么它在没有权力的情况下给了我答案，而有权力的人却没有给我答案。

Answer 1

number_of_months = 0
total_cost = 100000
semi_annual_income = 1000
starting_salary = 1000
semi_annual_rise = 0.2
number_of_months = 0
while semi_annual_income < total_cost:
    for i in range(1,100):
        if float(i)%6 == 0:
            power_number = float(i)/6# i am using this to make it increase the 
semi_annual income just only every six months
            number_of_months = number_of_months + 1
            semi_annual_income = starting_salary *(float(1) + 
float(semi_annual_rise))**(power_number)
            print(power_number)
            print(semi_annual_income)
            print(number_of_months)
# I think my mistake was the i used semi_annual-income instead of starting salary, this code works as i wanted it to be.
    break