Key Concepts and Formulas

To solve this problem, we need to utilize two fundamental concepts:

1. The Binomial Theorem: For an expansion of the form $(a+b)^n$, the general term, or the $(k+1)^{th}$ term, is given by T_{k+1} = \binom{n}{k} a^{n-k} b^k$$. In our specific problem, the expression is $$(1+y)^m$$. Therefore, the $$(k+1)^{th}$$ term is T_{k+1} = \binom{m}{k} (1)^{m-k} (y)^k = \binom{m}{k} y^k$. The coefficient of the$(k+1)^{th} term is $C_{k+1} = \binom{m}{k}. It is crucial to remember that the index 'k' in $\binom{m}{k}$ is one less than the term number.

2. Arithmetic Progression (AP): Three numbers, A, B, and C, are said to be in Arithmetic Progression if the difference between consecutive terms is constant. This implies that $B - A = C - B$, which simplifies to $2B = A + C$. This is the defining characteristic we will use.

Step-by-Step Solution

1. Identify the Coefficients of the Relevant Terms

We are given that the coefficients of the $r^{th}$, $(r+1)^{th}$, and $(r+2)^{th}$ terms are in A.P. First, we need to find the expressions for these coefficients using the binomial theorem.

• For the $r^{th}$ term: Since the $(k+1)^{th}$ term has coefficient $\binom{m}{k}$, for the $r^{th}$ term, we set $k+1 = r$. This means $k = r-1$. So, the coefficient of the $r^{th}$ term, denoted as $C_r$, is $\binom{m}{r-1}$.

• For the $(r+1)^{th}$ term: For the $(r+1)^{th}$ term, we set $k+1 = r+1$. This means $k = r$. So, the coefficient of the $(r+1)^{th}$ term, denoted as $C_{r+1}$, is $\binom{m}{r}$.

• For the $(r+2)^{th}$ term: For the $(r+2)^{th}$ term, we set $k+1 = r+2$. This means $k = r+1$. So, the coefficient of the $(r+2)^{th}$ term, denoted as $C_{r+2}$, is $\binom{m}{r+1}$.

2. Apply the Arithmetic Progression Condition

Since the coefficients $C_r$, $C_{r+1}$, and $C_{r+2}$ are in A.P., we use the condition $2B = A + C$ from the definition of A.P.: $2 \cdot C_{r+1} = C_r + C_{r+2}$ Substituting the binomial coefficients we found: $2 \binom{m}{r} = \binom{m}{r-1} + \binom{m}{r+1}$

3. Simplify the Equation Using Properties of Binomial Coefficients

To simplify this equation efficiently, we will use a common property of binomial coefficients that relates consecutive terms: $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$

Let's divide the entire AP equation by $\binom{m}{r}$ (assuming $\binom{m}{r} \neq 0$, which must be true for these terms to exist): $2 = \frac{\binom{m}{r-1}}{\binom{m}{r}} + \frac{\binom{m}{r+1}}{\binom{m}{r}}$

Now we simplify each ratio:

• For the first ratio, $\frac{\binom{m}{r-1}}{\binom{m}{r}}$: This is the reciprocal of the property above. If $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$, then $\frac{\binom{n}{k-1}}{\binom{n}{k}} = \frac{k}{n-k+1}$. Applying this with $n=m$ and $k=r$: $\frac{\binom{m}{r-1}}{\binom{m}{r}} = \frac{r}{m-r+1}$

• For the second ratio, $\frac{\binom{m}{r+1}}{\binom{m}{r}}$: Applying the property $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$ with $n=m$ and $k=r+1$ (since the numerator has index $r+1$ and the denominator has index $r$): $\frac{\binom{m}{r+1}}{\binom{m}{r}} = \frac{m-(r+1)+1}{r+1} = \frac{m-r}{r+1}$

Substitute these simplified ratios back into our AP equation: $2 = \frac{r}{m-r+1} + \frac{m-r}{r+1}$

4. Solve for the Relationship between m and r

Now, we need to perform algebraic manipulation to combine the terms and find the required equation. First, find a common denominator for the right-hand side, which is $(m-r+1)(r+1)$: $2 = \frac{r(r+1) + (m-r)(m-r+1)}{(m-r+1)(r+1)}$

Multiply both sides by the common denominator to eliminate fractions: $2(m-r+1)(r+1) = r(r+1) + (m-r)(m-r+1)$

Now, expand both sides carefully: $2(mr + m - r^2 - r + r + 1) = (r^2 + r) + (m^2 - mr + m - mr + r^2 - r)$ $2(mr + m - r^2 + 1) = r^2 + r + m^2 - 2mr + m + r^2 - r$ $2mr + 2m - 2r^2 + 2 = m^2 - 2mr + m + 2r^2$

Move all terms to one side to form a standard quadratic-like equation: $m^2 - 2mr + m + 2r^2 - (2mr + 2m - 2r^2 + 2) = 0$ $m^2 - 2mr + m + 2r^2 - 2mr - 2m + 2r^2 - 2 = 0$

Combine like terms: $m^2 + (-2mr - 2mr) + (m - 2m) + (2r^2 + 2r^2) - 2 = 0$ $m^2 - 4mr - m + 4r^2 - 2 = 0$

Finally, factor out 'm' from the terms containing 'm' to match the options provided: $m^2 - m(4r + 1) + 4r^2 - 2 = 0$

This equation represents the relationship between 'm' and 'r' for the given condition. Comparing this with the options, it matches option (C).

Tips for Success & Common Mistakes

• Correct Indexing: Always double-check that you are using the correct index $k$ for the $N^{th}$ term, which is $N-1$. A common mistake is to use 'r' for the $r^{th}$ term's coefficient as $\binom{m}{r}$, which is incorrect.
• Algebraic Precision: Binomial coefficient problems often involve extensive algebraic manipulation. Pay close attention to signs, distributive properties, and combining like terms. One small error can lead to a completely different result.
• Leverage Identities: Memorizing and correctly applying binomial coefficient identities (like the ratio identity used here) can significantly simplify complex algebraic expressions and save time compared to expanding factorials directly.
• Contextual Constraints: Remember that for binomial coefficients $\binom{n}{k}$ to be meaningful, $n \ge k$ and $k \ge 0$. In this problem, this implies $m \ge r+1$ and $r-1 \ge 0$ (so $r \ge 1$).

Summary and Key Takeaway

This problem seamlessly integrates your understanding of the Binomial Theorem with the properties of Arithmetic Progression. The most efficient path to the solution involves correctly identifying the coefficients for the specified terms, setting up the AP condition ($2B = A+C$), and then skillfully simplifying the resulting equation using the ratio property of binomial coefficients ($\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$). This method avoids the more cumbersome expansion of factorials. The final derived relationship, $m^2 - m(4r + 1) + 4r^2 - 2 = 0$, highlights how these mathematical concepts interrelate.