2
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ $x_2=x_{1+1}=\dfrac{x_1}{1}$ $\endgroup$ – Martigan Jun 4 '15 at 15:10
5
$\begingroup$

First step is to find the recurrence:

$$\begin{split} x_{n+1} &= \frac{\sum_{i=1}^nix_i}n \\ &=\frac{\sum_{i=1}^{n-1}ix_i + nx_n}n \\ &=\frac{(n-1)\frac{\sum_{i=1}^{n-1}ix_i}{n-1} + nx_n}n \\ &=\frac{(n-1)x_n + nx_n}n \\ &= \frac{2n-1}nx_n \end{split}$$

$\endgroup$
2
$\begingroup$

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 $$x_{n+1} =x_1 \cdot\prod_{j=1}^n \frac{x_{j+1}}{x_j} =x_1\cdot\prod_{j=1}^n \left(2-\frac{1}{j}\right).$$

$\endgroup$
2
$\begingroup$

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 $$ \begin{align} x_{n+1} &=\frac{(2n-1)!!}{n!}x_1\\[6pt] &=\frac{a}{2^n}\binom{2n}{n}\tag{4} \end{align} $$

$\endgroup$
0
$\begingroup$

Hint: $x_{n+1}=\dfrac{2n-1}{n}x_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.