# Find $x_{n}$ if $x_{1}=a>0$ and $x_{n+1}=\frac{x_{1}+2x_{2}+…+nx_{n}}{n}$

Find $x_{n}$ if $x_{1}=a>0$ and $x_{n+1}=\frac{x_{1}+2x_{2}+...+nx_{n}}{n}$

I have a problem finding sum of $$x_{1}+2x_{2}+...+nx_{n}$$ I don't see the term $x_{2}$ because if $x_{1}=a$ for $n=1$, then for $n=2$ it would be $x_{3}=\frac{x_{1}+2x_{2}}{2}$

How to determine the sum?

• $x_2=x_{1+1}=\dfrac{x_1}{1}$ – Martigan Jun 4 '15 at 15:10

$$\begin{split} x_{n+1} &= \frac{\sum_{i=1}^nix_i}n \\ &=\frac{\sum_{i=1}^{n-1}ix_i + nx_n}n \\ &=\frac{(n-1)\frac{\sum_{i=1}^{n-1}ix_i}{n-1} + nx_n}n \\ &=\frac{(n-1)x_n + nx_n}n \\ &= \frac{2n-1}nx_n \end{split}$$
We have $$nx_{n+1} - (n-1) x_n = (x_1 +2x_2 +...+ nx_n ) -(x_1 +2x_2 +...+ (n-1)x_{n 1} ) =nx_n$$ hence $$nx_{n+1} =(2n-1)x_n$$ thus $$\frac{x_{n+1}}{x_n} =\left( 2-\frac{1}{n} \right)$$ So $$x_{n+1} =x_1 \cdot\prod_{j=1}^n \frac{x_{j+1}}{x_j} =x_1\cdot\prod_{j=1}^n \left(2-\frac{1}{j}\right).$$
We have $$nx_{n+1}=\sum_{k=1}^nkx_k\tag{1}$$ and substituting $n\mapsto n-1$, $$(n-1)x_n=\sum_{k=1}^{n-1}kx_k\tag{2}$$ Subtract $(2)$ from $(1)$ to get $nx_{n+1}-(n-1)x_n=nx_n$ which is the same as $$x_{n+1}=\frac{2n-1}{n}x_n\tag{3}$$ Therefore, induction says \begin{align} x_{n+1} &=\frac{(2n-1)!!}{n!}x_1\\[6pt] &=\frac{a}{2^n}\binom{2n}{n}\tag{4} \end{align}
Hint: $x_{n+1}=\dfrac{2n-1}{n}x_n$.