具体数学--(求和式的一般形式的几种方法)

本文链接：https://blog.youkuaiyun.com/luixiao1220/article/details/103718572

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How to find the closed form of any formula? We will demonstrate an
example to show seven different methods to achieve this goal.

$S_n=\sum_{1\leqslant k\leqslant n}k^2,n\geqslant 0$

You could look it up.

There’s several books that you can look it up.

Standard Mathematical Tables
Handbook of Mathematical Functions
Handbook of Integer Sequences

I think google is the first choice of the modern people.

Guess the answer, prove it by induction

It is very difficult to guess the answer, so this method is not
recommended.

Perturb the sum

First method

$S_n=\sum_{i=1}^{n}i^2$

Perturb $\left. \begin{aligned} S_n+(n+1)^2&=\sum_{0\leqslant k\leqslant n}(k+1)^2=\sum_{0\leqslant k\leqslant n}k^2+2\sum_{0\leqslant k\leqslant n}k+\sum_{0\leqslant k\leqslant n} 1\\ &=S_n+2 \sum_{0\leqslant k\leqslant n}k+(n+1) \end{aligned} \right.$

Finally we can get that
$S_n+(n+1)^2=S_n+2 \sum_{0\leqslant k\leqslant n}k+(n+1)$

The left $S_n$ and right $S_n$ cancelled each other which leads to a
failure of our process. However, we get the solution of
$(n+1)^2=2 \sum_{0\leqslant k\leqslant n}k+(n+1) \Rightarrow \sum_{0\leqslant k\leqslant n}k=[(n+1)^2-(n+1)]/2$

This clue leads to a second method which begin with $i^3$ .
Second Method

$T_n=\sum_{i=1}^{n}i^3$

Perturb

$\left. \begin{aligned} T_n+(n+1)^3&=\sum_{0\leqslant k\leqslant n}(k+1)^3=\sum_{0\leqslant k\leqslant n} k^3 + 3\sum_{0\leqslant k\leqslant n}k^2+3\sum_{0\leqslant k\leqslant n}k+\sum_{0\leqslant k\leqslant }1\\ &=T_n+3S_n+3 \frac{(n+1)n}{2}+(n+1) \end{aligned} \right.$

Finally, we get

$3S_n=(n+1)(n+\frac{1}{2})n$

Build a repertoire

This is a general method. We can assume that the solution to be

$\left. \begin{aligned} R_0&=\alpha\\ R_n&=R_{n-1}+\beta+\gamma n+ \delta n^2,n>0 \end{aligned} \right.$

Now, we have four parameters $\alpha,\beta, \gamma, \delta$ . The closed
form should be

$R_n=A(n)\alpha+B(n)\beta+C(n)\gamma+D(n)\delta$

At this step, our focus is on the $A (n), B (n), C (n), D (n)$ . We will use
special $\alpha,\beta,\gamma,\delta$ to get these
$A (n), B (n), C (n), D (n)$ .

$A (n)$
1. Let’s assume that $\alpha=1,\beta=0,\gamma=0,\delta=0$
  $R_n=R_{n-1}\Rightarrow R_n=R_{n-1}=...=R_0=1(R_0=\alpha=1)$
2. Now, we get $\mathbf{A(n)=1}$
$B (n)$
1. $R_n=f(n)=n$
2. $n=n-1+\beta+\gamma n+\delta n^2 \Rightarrow 1=\beta+\gamma n+\delta n^2 \Rightarrow \beta=1,\gamma=0,\delta=0$
3. $f(n)=A(n)\alpha+B(n)\beta \Rightarrow f(n)=1 \cdot \alpha+B(n) 1 \Rightarrow f(n)=R_n=n=\alpha+B(n)$
4. $\mathbf{B(n)=n}$
$C (n)$
1. $R_n=f(n)=n^2$
2. $n^2=(n-1)^2+\beta+\gamma n+\delta n^2 \Rightarrow 2n-1=\beta+\gamma n+\delta n^2 \Rightarrow \beta=-1,\gamma=2,\delta=0, \alpha=0$
3. $f(n)=n^2=2C(n)-n \Rightarrow \mathbf{C(n)=\frac{n^2+n}{2}}$
$D (n)$
1. $R_n=f(n)=n^3 \Rightarrow \mathbf{3D(n)=n(n+\frac{1}{2})(n+1)}$

Finally, we get what we want.

$A(n)=1,B(n)=n,C(n)=\frac{n+n^2}{2},D(n)=\frac{n(n+\frac{1}{2})(n+1)}{3}$

After we know $A (n), B (n), C (n), D (n)$ , we can infer
$\alpha,\beta,\gamma,\delta$ for specific problems such as
$R_0=0,R_n=R_{n-1}+n^2$ .

$R_0=0 \Rightarrow R_0=\alpha=0$
$R=R_{n-1}+n^2 \Rightarrow \beta=0,\gamma=0,\delta=1$

At last, we get

$R(n)=\frac{n(n+\frac{1}{2})(n+1)}{3}$

Replace sums by integrals

$\int_{0}^{n}{x^2dx}=\frac{1}{3}x^3|_{0}^{n}=\frac{1}{3}n^3$

$S_n=\sum_{i=0}^{n}k^2$

How to calculate $S_n$ based on integral?

$E_n=S_n-\frac{1}{3}n^3= S_{n-1}+n^2-\frac{1}{3}n^3,(S_{n-1}=E_{n-1}+\frac{1}{3}(n-1)^3)$
$E_n=E_{n-1}+\frac{1}{3}(n-1)^3+n^2-\frac{1}{3}n^3=E_{n-1}+n-\frac{1}{3}$

From $E_n=E_{n-1}+n-\frac{1}{3}$ , we can get a closed form for $E_n$ ,
and when we get the closed form for $E_n$ , we get the closed form for
$S_n$

Another path

$E_n=S_n-\int_{0}^{n}{x^2dx}=\sum_{k=1}^{n}(k^2-\int_{k-1}^{k}{x^2dx})=\sum_{k=1}^{n}(k^2-\frac{k^{3-(k-1)^3}}{2})$
$E_n=\sum_{k=1}^{n}(k-\frac{1}{3})$

The same result as previous one.

Convert multiple to sum

$\left. \begin{aligned} S_n&=\sum_{1\leqslant k\leqslant n}k^2=\sum_{1\leqslant j\leqslant k\leqslant n}k\\ &=\sum_{1\leqslant j\leqslant n}\sum_{j\leqslant k\leqslant n}k\\ &=\sum_{1\leqslant j\leqslant n}(\frac{j+n}{2})(n-j+1)\\ &=\frac{1}{2}\sum_{1\leqslant j\leqslant n}(n(n+1)+j-j^2)\\ &=\frac{1}{2}n^2(n+1)+\frac{1}{4}n(n+1)-\frac{1}{2}S_n=\frac{1}{2}n(n+\frac{1}{2})(n+1)-\frac{1}{2}S_n \end{aligned} \right.$