Animated Solution for Mathematics - Binomial Theorem: Using induction or otherwise, prove that for any non-negative integers $m,n,r$ and $k$, ${\sum }_{m=0}^k(n-m)\frac{(r+m)!}{m!}=\frac{(r+k+1)!}{k!}[\frac{n}{r+1}-\frac{k}{r+2}]$

Welcome, fellow JEE warriors. Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of mathematical induction.

When you first look at the expression ${\sum }_{m=0}^k(n-m)\frac{(r+m)!}{m!}$, it is natural to feel a surge of intimidation. It is a sum of factorials, a structure that often hides deep complexity.

But remember, in the world of JEE Advanced, intimidation is just a signal that you are about to learn something beautiful. Our mission is to prove that this sum simplifies to:

The Base Case

Every great structure needs a solid foundation. In induction, this is our base case, $k=0$.

When we set $k=0$, the summation collapses into a single term where $m=0$. The left-hand side becomes $(n-0)\frac{(r+0)!}{0!}$, which simplifies to $n\cdot r!$.

Now, we check the right-hand side. Substituting $k=0$ into the formula gives us:

Since $(r+1)!=(r+1)\cdot r!$, the $(r+1)$ terms cancel out, leaving us with $n\cdot r!$. The base case holds! The first domino has fallen.

The Inductive Hypothesis

Now, we move to the heart of the proof. We assume the identity holds for an arbitrary integer $k$.

This is our Inductive Hypothesis:

Treat this as a given truth. It is the magic key that will unlock the next step. We are not just guessing; we are building a bridge to the next integer.

The Inductive Step

This is where the real work happens. We want to find the sum up to $k+1$, which we call $S_{k+1}$.

Instead of starting from scratch, we use the recursive nature of the sum: $S_{k+1}=S_k+T_{k+1}$, where $T_{k+1}$ is the $(k+1)$-th term. Substituting our hypothesis for $S_k$ and adding the $(k+1)$-th term, we get a long, complex expression.

Do not panic. This is where the algebra becomes an art form. We notice that both parts of our expression contain $(r+k+1)!$. By adjusting the denominators to match, we can factor out $\frac{(r+k+1)!}{(k+1)!}$.

Inside the brackets, we are left with:

Algebraic Harmony

Now, we focus on the expression inside the brackets. We distribute the $(k+1)$ and group the terms containing $n$ and the terms that do not.

This is a standard, powerful maneuver. After grouping, we find that both parts share a common factor of $(k+r+2)$.

When we factor this out, the expression inside the bracket becomes:

Finally, we combine the $(k+r+2)$ with the factorial term outside. Since $(r+k+1)!\cdot (k+r+2)=(r+k+2)!$, our expression transforms into:

This is exactly the original formula with $k$ replaced by $k+1$. The proof is complete. The dominoes have fallen in perfect sequence.

Remember, math is not about memorizing steps; it is about seeing the patterns and trusting the logic. You have just conquered a complex identity. Keep practicing, keep questioning, and most importantly, keep falling in love with the process.