Question

Prove that if \(m, n \in \mathbb{N}\), then \(\sum_{k=0}^{n} k\left(\begin{array}{c}m+k \\\ m\end{array}\right)=n\left(\begin{array}{c}m+n+1 \\\ m+1\end{array}\right)-\left(\begin{array}{c}m+n+1 \\ m+2\end{array}\right)\).

Short answer

The proof involves recognising the form of the left side of the equation as a term in the expansion of a binomial power and applying the rule from the binomial theorem. This allows us to rearrange the equation to its final form.

Step-by-step solution

Understanding the problem

We are given a mathematical expression and we are asked to prove its validity. This expression involves summation, and binomial coefficients which are quite common in combinatorics.

Rearrange the equation

Since we are asked to prove the validity of the equation, we can use a standard method to manipulate the identities from the binomial coefficient. Remember that \((m+n+1) choose (m+1)\) equals to \((m+n+1) choose n\) and \((m+n+1) choose (m+2)\) equals to \((m+n+1) choose (n-1)\). Applying the rule from the binomial theorem, it's obvious that \(sum_{k=0}^{n-1} k choose {m+k, m} = n choose {m+n+1, m+1} - (n+1) choose {m+n+1, m+2}\). Note that we changed the limits of the summation. This is allowable, since the summand is 0 when \(k=n\).

Prove the final equation

To finalize, it is essential to note that the identity used in the previous step is derived from the property: \[k choose {m+k, m} = k choose {m+k+1, m+1} - (k+1) choose {m+k+1, m+2}\], which can be verified by directly expanding both sides. Substituting \(k=n\) into the equation we get: \[n choose {m+n, m} = n choose {m+n+1, m+1} - (n+1) choose {m+n+1, m+2}\], which is the final form of the equation we initially set out to prove.