这是一份southampton南安普敦大学MATH2014W1-01作业代写的成功案例
$$
x_{n+1}=3 x_{n}+n \quad\left(n \geq 0 ; x_{0}=0\right) .
$$
The winning change of variable, is to let $x_{n}=3^{n} y_{n}$. After substituting and simplifying, we find
$$
y_{n+1}=y_{n}+n / 3^{n+1} \quad\left(n \geq 0 ; y_{n}=0\right)
$$
Now by summation,
$$
y_{n}=\sum_{j=1}^{n-1} j / 3^{j+1} \quad(n \geq 0) .
$$
Finally, since $x_{n}=3^{n} y_{n}$ we obtain in the form
$$
x_{n}=3^{n} \sum_{j=1}^{n-1} j / 3^{j+1} \quad(n \geq 0) .
$$
MATH2014W1-01 COURSE NOTES ：
The quantities $n ! / k !(n-k) !$ are the famous binomial coefficients, and they are denoted by
$$
\left(\begin{array}{l}
n \
k
\end{array}\right)=\frac{n !}{k !(n-k) !} \quad(n \geq 0 ; 0 \leq k \leq n)
$$
Some of their special values are
$$
\begin{gathered}
\left(\begin{array}{l}
n \
0
\end{array}\right)=1 \quad(\forall n \geq 0) ; \quad\left(\begin{array}{l}
n \
1
\end{array}\right)=n \quad(\forall n \geq 0) ; \
\left(\begin{array}{l}
n \
2
\end{array}\right)=n(n-1) / 2 \quad(\forall n \geq 0) ; \quad\left(\begin{array}{l}
n \
n
\end{array}\right)=1 \quad(\forall n \geq 0) .
\end{gathered}
$$
It is convenient to define $\left(\begin{array}{l}n \ k\end{array}\right)$ to be 0 if $k<0$ or if $k>n$.