Question

If \(n\) and \(k\) are non-negative integers, then \(\left(\begin{array}{c}n+0 \\\ 0\end{array}\right)+\left(\begin{array}{c}n+1 \\\ 1\end{array}\right)+\left(\begin{array}{c}n+2 \\\ 2\end{array}\right)+\cdots+\left(\begin{array}{c}n+k \\\ k\end{array}\right)=\left(\begin{array}{c}n+k+1 \\ k\end{array}\right)\).

Short answer

The identity \(\left(\begin{array}{c}n \ 0\end{array}\right)+\left(\begin{array}{c}n+1 \ 1\end{array}\right)+\left(\begin{array}{c}n+2 \ 2\end{array}\right)+\cdots+\left(\begin{array}{c}n+k \ k\end{array}\right)=\left(\begin{array}{c}n+k+1 \ k\end{array}\right)\) holds true for all non-negative integers \(n\) and \(k\), as proven by mathematical induction.

Step-by-step solution

Base Case

We need to establish the base case for \(k=0\). The left side of the identity becomes \(\left(\begin{array}{c}n \ 0\end{array}\right)\) which equals 1 according to the definition of binomial coefficients. The right side of the identity for \(k=0\) is \(\left(\begin{array}{c}n+1 \ 0\end{array}\right)\) which also equal 1. Hence, the base case holds true.

Inductive Hypothesis

We will assume that the identity holds for some \(k=m\), i.e., \(\left(\begin{array}{c}n \ 0\end{array}\right)+\left(\begin{array}{c}n+1 \ 1\end{array}\right)+\left(\begin{array}{c}n+2 \ 2\end{array}\right)+\cdots+\left(\begin{array}{c}n+m \ m\end{array}\right)=\left(\begin{array}{c}n+m+1 \ m\end{array}\right)\). This is the induction hypothesis.

Inductive Step

Now we need to prove that the identity holds for \(k=m+1\) if it holds for \(k=m\). According to the identity, the left side for \(k=m+1\) is \(\left(\begin{array}{c}n \ 0\end{array}\right)+\left(\begin{array}{c}n+1 \ 1\end{array}\right)+\left(\begin{array}{c}n+2 \ 2\end{array}\right)+\cdots+\left(\begin{array}{c}n+m+1 \ m+1\end{array}\right)\). We can separate this into the sum for \(k=m\), and the term for \(k=m+1\), i.e., \(\left(\begin{array}{c}n+m+1 \ m\end{array}\right) + \left(\begin{array}{c}n+m+1 \ m+1\end{array}\right)\). We can use the binomial identity \(\left(\begin{array}{c} n \ r \end{array}\right) + \left(\begin{array}{c} n \ r+1 \end{array}\right) = \left(\begin{array}{c} n+1 \ r+1 \end{array}\right)\) to combine these two terms to get \(\left(\begin{array}{c}n+m+2 \ m+1\end{array}\right)\), which is the right side of the identity for \(k=m+1\). Thus, the inductive step is verified.