Let $n$ be a positive integer. Assume that:
$N_k$ is the number of pairs $(a,b)$ of non-negative integers such that $ka+(k+1)b=n+1-k$. Find $N_1+N_2+\cdots N_{n+1}$.
I was trying to solve this calculating coefficients of polynomial equations but in vain like:
We know that $N_k$ is the coefficient of $X^{n-k+1}$ in:
$$G_k(X)=(1+X^k+X^{2k}+\cdots)(1+X^{k+1}+X^{2k+2}+\cdots)$$
Now can you help me from here or by any other elegant way. Please help.
Set $N=N_1+\dots+N_{n+1}$ and notice that $N$ is precisely the number of integer solutions $(k,a,b)$ to $ka+(k+1)b=n+1-k$ where $k\geq 1$ and $a,b\geq 0$. Setting $q=a+b+1$ and $r=b$ and rearranging the expression, $N$ is the number of integer solutions $(k,q,r)$ to $n+1=qk+r$ where $k\geq 1$ and $0\leq r\leq q-1$. But this expression is exactly what you get when dividing $n+1$ by $q$. More specifically, $k$ is the quotient, and $r$ is the remainder. Thus, we know that for every fixed $1\leq q\leq n+1$, there is precisely one integer solutions $(k,q,r)$ to $n+1=qk+r$ where $k\geq 1$ and $0\leq r\leq q-1$, so $N=n+1$.