$N_1$ and $N_2$ are cyclic submodules Please help me to prove the following result:

Let $R$ be a ring with $1$ and $M$ a noetherian  left $R$-module. Suppose that if $N_1$ and $N_2$ are cyclic submodules of $M$ than $N_1+N_2$ is also cyclic.


  Show that $M$ is cyclic 


Thanks
An often useful equivalent property to the ascending chain condition is the maximal condition:


  the module $M$ is noetherian if and only if every nonempty set of submodules of $M$ has a maximal element


(maximality is with respect to set inclusion.

Since $M$ is noetherian, the set of its cyclic submodules has a maximal element $N$; since $\{0\}$ is cyclic, this set is not empty. If $N\ne M$, then take $x\in M$, $x\notin N$. By assumption, $N+xR$ is a cyclic submodule, where $xR$ denotes the cyclic submodule generated by $x$: a contradiction to maximality, because $N+xR\supsetneq N$ as $x\in N+xR$. Hence $N=M$.
@ Tobias Kildetoft , Is it right if i answer like the following:

Let $m_0\in M$ and assume that $m_0$ does not generate $M$. We pick an $x_1$ not in the submodule generated by $m_0$ and let $m_1$ be a generator for the submodule generated by $m_0$ and $x_1$. We repeat the argument to get larger and larger submodules. We are picking elements not in the submodule generated by various elements in order to get the next one and  we are using the fact the the sum is cyclic to pick a generator at each step. Thus we obtain an ascending chain
$$(m_0)⊂(m_1)⊂(m_2)⊂....$$
of submodules of $M$. Since $M$ is noetherian there exists some integer $n_0$ such that $(m_n)=(m_{n+1})$ for all $n\ge n_0$.  This means we have $M=(m_{n_0})$ and thus $M$ is cyclic.