從n的一次方加到n的五十次方是多少

這是用了Geometric series

要求n+n^2+...+n^50, 一定要用G.P.公式(form 5 Maths)


Geometric series explanation:

A geometric series is one in which there is a constant ratio between each element and the one preceding it. Here is one such series.

7+14+28+56+112

The ratio here is 2. Let's try to find the sum of geometric series, in general (S is the sum, a is the first element, r is the ratio, n is the number of elements):

S=a+ar+ar^2+ar^3+...+ar^(n-1)
Sr=a^r+ar^2+ar^3+...+ar^(n-1)+ar^n
S-Sr=a-a(r^n) [subtracting the second line from the first]
S(1-r)=a(1-r^n)
S=a(1-r^n)/(1-r)

這就是公式

如題，Ans: S=n*(1-n^50)/(1-n)

Hope this is useful to you!

2007-01-11 17:59:10 補充：
其實我的答案和n( n^50- 1) / (n-1) 一樣，只不過上下乘了負1。兩個答案均可接受。

2007-01-12 13:28:52 補充：
下面唔好抄我！ ={

n^1+n^2+n^3+...+n^50
=(n^51-n)/(n-1)

這其實是用了Geometric series

要求n+n^2+...+n^50, 一定要用G.P.公式(form 5 Maths)


Geometric series explanation:

A geometric series is one in which there is a constant ratio between each element and the one preceding it. Here is one such series.

7+14+28+56+112

The ratio here is 2. Let's try to find the sum of geometric series, in general (S is the sum, a is the first element, r is the ratio, n is the number of elements):

S=a+ar+ar^2+ar^3+...+ar^(n-1)
Sr=a^r+ar^2+ar^3+...+ar^(n-1)+ar^n
S-Sr=a-a(r^n) [subtracting the second line from the first]
S(1-r)=a(1-r^n)
S=a(1-r^n)/(1-r)

這就是公式

如題，Ans: S=n*(1-n^50)/(1-n)

Hope this is useful to you!

希望你參考我的意見啦!Thank you!

可以利用等比級數公式~
n(n^50)
------------
n-1
=n^51/(n-1)

2007-01-11 17:57:45 補充：
n(n^50-1)------------n-1=n(n^50-1)/(n-1)

2007-01-11 18:00:32 補充：
等比級數公式是初值x(公比^項數-1)-----------------------------公比-1

n( n^50- 1) / (n-1)