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系列 1 + (1+2+1) + (1+2+1+3+1+2+1) 的总和

[英]Summation of series 1 + (1+2+1) + (1+2+1+3+1+2+1)

The function is: F(n-1) n F(n-1) function 为:F(n-1) n F(n-1)

Its a type of palindrome function called Zimmer Series.它是一种称为 Zimmer 系列的回文 function。

The values would be: 1, 121, 1213121, ...值将是:1、121、1213121、...

I want to figure the summation of the individual digits.我想计算各个数字的总和。

1 + (1+2+1) + (1+2+1+3+1+2+1) +... 1 + (1+2+1) + (1+2+1+3+1+2+1) +...

Any help is welcome.欢迎任何帮助。

Breaking this down into steps, we first find out a formula for the summation of a single value of the series and then we can find out the summation of said formula.将其分解为步骤,我们首先找出一个公式,用于对序列的单个值求和,然后我们可以找出该公式的求和。

Expanding the definition you gave and manipulating it:扩展您给出的定义并对其进行操作:

F(n) = n + 2F(n-1)
F(n) = n + 2(n-1) + 22(n-2) + 23(n-3) + ... + 2n-1
2F(n) =    2n     + 22(n-1) + 23(n-2) + ... + 2n-1(2) + 2n

F(n) - 2F(n) = -F(n) = n - 2 - 22 - 23 - ... - 2n

From this and using the formula for Geometric Progression we can then get an expression for a single term of the series.由此并使用几何级数的公式,我们可以得到该系列的单个术语的表达式。

F(n) = (2n + 2n-1 + ... + 2) - n
     = (2n+1 - 2) - n

Now we just have to work out the summation of this expression.现在我们只需要计算出这个表达式的总和。

G(n) = Σ F(n) = Σ (2n+1 - 2 - n)
G(n) = (2n+2 - 22) - (2n) - (n(n+1)/2)

Simplifying this should hopefully give you the answer you seek!简化这应该有望给你你寻求的答案!

G(n) = (2n+2  - (n(n+5)/2) - 22)

Trying this out on a few of the terms just to double check.在一些条款上尝试这个只是为了仔细检查。

G(1) = (21+2 - (1(1+5)/2) - 22)
G(1) = 1
G(2) = (22+2  - (2(2+5)/2) - 22)
G(2) = 5 = 1 + (1 + 2 + 1)
G(3) = (23+2  - (3(3+5)/2) - 22)
G(3) = 16 = 1 + (1 + 2 + 1) + (1 + 2 + 1 + 3 + 1 + 2 + 1)

EDIT: Mark Dickinson is right, I misinterpreted the question, this solution is incorrect.编辑:马克狄金森是对的,我误解了这个问题,这个解决方案是不正确的。

I think after the second term the sequence is of the form difference in Arithmetic Progression.我认为在第二学期之后,该序列在算术级数中具有形式差异。

Let me show you how让我告诉你怎么做

Second Term = 1+2+1
Third Term  = 1+2+1+3 + 1+2+1
Difference  = 1+2+1+3 = 7 

Third Term  = 1+2+1+3+1+2+1
Fourth Term = 1+2+1+3+  1+4+1+3 +1+2+1
Difference  = 1+4+1+3 = 9

Fourth Term = 1+2+1+3+1+4+1+3+1+2+1
Fifth Term  = 1+2+1+3+1+4+  1+5+1+4  +1+3+1+2+1
Difference  = 1+5+1+4 = 11

So as you can see the difference is in the arithmetic progression and you find the sum of the terms using the formula for the sum of numbers whose different are in Arithmetic Progression因此,您可以看到差异在于算术级数,您可以使用算术级数中不同的数字之和的公式找到各项的总和

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