<s>[INST] Find $F_{n}$ in : $F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} = 3^{n}$ I'm stuck with the question for a while : Find in $$F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} = 3^{n}$$
the element $F_{n}$ .
Placing $n-1$ instead on $n$ results in : 
$$F_{n-1} +2F_{n-2} + ... + (n-1+1)\cdot F_{0} = 3^{n-1}$$
$$ F_{n-1} +2F_{n-2} + ... + n\cdot F_{0} = 3^{n-1} $$
subtracting both : 
$$F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} -(F_{n-1} +2F_{n-2} + ... + n\cdot F_{0} )
=3^{n} - 3^{n-1}
$$
But that doesn't help much . Any ideas ? 
Thanks 
 [/INST] The condition
$$F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} = 3^{n}$$
is equivalent to the equality of coefficients of $z^n$ in below expression:
$$( F_0 + F_1 z + F_2 z^2 + \cdots )(1 + 2z + 3z^2 + \cdots ) = 1 + 3z + 3^2 z^2 + \cdots$$
Notice $\frac{1}{(1-z)^2} = 1 + 2z + 3z + \cdots$ and $\frac{1}{1-3z} = 1 + 3z + 3^2 z^2 + \cdots$, we have:
$$\frac{\sum_{k=0}^{\infty} F_k z^k}{(1-z)^2} = \frac{1}{1-3z} \quad\implies\quad\sum_{k=0}^{\infty} F_k z^k  = \frac{(1-z)^2}{1-3z} = \frac{1 - 2z + z^2}{1-3z}$$
By comparing the coefficients of $z^n$, we get:
$$\begin{align}
F_0 &= 1\\
F_1 &= 1\cdot 3^1 - 2 \cdot 1 = 1\\
&\;\vdots\\
F_n &= 1\cdot 3^n - 2 \cdot 3^{n-1} + 1 \cdot 3^{n-2} = 4\cdot 3^{n-2},\quad\text{ for }n \ge 2
\end{align}$$
</s>