Prove that for fixed positive integers $k$ and $n$, the number of partitions of $n$ is equal to the number of partitions of $2n + k$ into $n + k$ parts.
I tried this with different values of n but can't seem to get a bijection between the two partitions.
let $n=a_1+a_2+ \cdots +a_j$ be a partition of $n$ then \begin{eqnarray*} 2n+k= (a_1+1)+(a_2+1)+\cdots +(a_j+1) + \underbrace{1+1+ \cdots +1}_{\text{$n+k-j$ ones}}. \end{eqnarray*}