Question

如何使用投资5000的几何序列，每年增加5％的新值？我一直在尝试一切，但我无法得到正确答案。请帮忙

所以今年： 0 = 5000

5250
5512.5
5788.13

代码：

 public IActionResult Show(string generic_se_name)
    {
        var type = db.UrlRecords.FirstOrDefault(x => x.Slug == generic_se_name);
        if (type != null)
        {
            switch (type.EntityType)
            {
                case "Product":
                    return RedirectToRoute("Product", new {productName = generic_se_name });
                case "Category":
                    return RedirectToRoute("Category", new { categoryName = generic_se_name });
            }
        }
        return BadRequest();
    }

Answer 1

问题在于您的打印声明。

你有两个选择让它运作：

列表理解，创建完整的利润列表，然后用年数拉链

def calc_profits(principal, year): INTREST_GROWTH = 0.05 profits = [principal * ((1+INTREST_GROWTH) ** (y - 1)) for y in range(1, year + 1)] for i,p in zip(range(year),profits): print("{0}:{1:.2f}".format(i, p))
迭代给出一个新值并为每次循环迭代打印

def calc_profits(principal, year): INTREST_GROWTH = 0.05 profits = principal for y in range(1, year): profits = profits*(INTREST_GROWTH+1) print("{0}:{1:.2f}".format(y, profits))

Answer 2

您可以计算投资的最终价值，并拿走本金以了解您获得了多少利润。

def calc_profits(principal, year):
    INTEREST_GROWTH = 0.05
    end_capital = principal * (1 + INTEREST_GROWTH) ** year
    profit = end_capital - principal
    return profit

编辑：如果你想知道每年的利润，你可以用列表推导来做。

def calc_profits(principal, year):
    INTEREST_GROWTH = 0.05
    profit = [principal * (1 + INTEREST_GROWTH) ** y - principal
               for y in range(1, year + 1)]
    return profit

python中的几何序列

2 个答案: