let $a_{n}=n(a_{1}+a_{2}+\dots+a_{n-1})$, $n\geqslant 2$, $a_{1}=1$. Find $a_{n}$ let $a_{n}=n(a_{1}+a_{2}+\dots+a_{n-1})$, $n\geqslant 2$, $a_{1}=1$. Find $a_{n}$
My Approach:
so I thought, since this is a homogeneous relation, I expanded the expression and wrote the characteristic equation of the same, since that was $n(x^1+x^2+\dots+x^{n-1})$. So I substituted the values of $n=2$, and my answer was $n(n-1)$ which is wrong.
EDIT: I cannot use the characteristic equation since the coeff is not constant
 A: We have
\begin{align}
&{a_n} = n\left(\sum_{i=1}^{n-1}a_i\right)\\
\implies &{a_n}=n\left(\frac{a_{n-1}}{n-1}+a_{n-1}\right)\\
\implies &\frac{a_n}{n} = \left(\frac{a_{n-1}}{n-1}\right)\times n\\
\implies &\frac{a_n}{n\cdot n!}=\frac{a_{n-1}}{(n-1)\cdot(n-1)!}=\dots=\frac{a_2}{2\cdot 2!}=\frac12\\
\implies &\boxed{a_n = \frac{n\cdot n!}2}\qquad\forall n\ge 2
\end{align}
A: With $S_n = \sum_{k=1}^n a_k$ we have
$$
a_n = n S_{n-1} = n(S_n-a_n) = nS_n - n a_n\Rightarrow (n+1)a_n = n S_n
$$
so
$$
a_n = \frac{n}{n+1}S_n = n S_{n-1}
$$
and finally
$$
S_n = (n+1)S_{n-1},\ \ S_1 = 1
$$
and
$$
a_n = n\frac{n!}{2}
$$
A: We have $$\frac{a_n}{n} = a_1+a_2+\cdots +a_{n-1}= (a_1+a_2+\cdots +a_{n-2}) + a_{n-1} = \frac{a_{n-1}}{n-1}+a_{n-1} = n\cdot \frac{a_{n-1}}{n-1}$$
which is valid for $n\ge 3$ (notice that it's not valid for $n=2$ because we used $a_{n-2}$).
Then $$\frac{a_n}{n}= n\cdot \frac{a_{n-1}}{n-1}=n(n-1)\cdot \frac{a_{n-2}}{n-2}=\cdots = n(n-1)\cdots 3\cdot \frac{a_{2}}{2} = \frac{n!}{2}$$
A: $$a_1=1$$
$$a_2=2(a_1)=2$$
$$a_3=3(a_2+a_1)=9$$
$$a_4=4(a_1+a_2+a_3)=48$$
$$a_5=5(a_1+a_2+a_3+a_4)=300$$
Plugging this into OEIS gives A074143, where Bakare Gatta Naimat indicates that it maintains the general rule for $n\geq 2$ of $a_n=\frac{n(n!)}{2}$.
A: a(n) = n(a1+a2+a3....an-1)
So,
a(n+1) = (n+1) ( a1+a2....an) .
But a1+a2+...an-1=a(n)/n , from previous statement.
a(n+1) = (n+1)((a(n)/n )  +   an)
Rearranging we get:
a(n+1)/a(n) = (n+1)^2/n
On repeated multiplication starting at n=2 under the constraint that a(2)=2, we get,
a(n+1) = (n+1)*(n+1)fac/2
Which implies,
a(n) = (n)(nfac)/2
This is a general strategy used in sequence , in which we relate a(n+1) and a(n) using definition of the sequence and in most cases we get a telescoping sum or product . In this case a telescoping product.
