Key Concept: Binomial Theorem and Coefficients

The Binomial Theorem states that for any positive integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{r=0}^{n} {}^n C_r a^{n-r} b^r$ For the expansion of $(1+x)^n$, where $a=1$ and $b=x$, the general term (or the coefficient of $x^r$) is given by: $T_{r+1} = {}^n C_r (1)^{n-r} (x)^r = {}^n C_r x^r$ Therefore, the coefficient of $x^r$ in the expansion of $(1+x)^n$ is ${}^n C_r$.

Step 1: Identify the Coefficients

We need to find the coefficients of $x^4, x^5$, and $x^6$. Using the general formula for the coefficient of $x^r$:

• Coefficient of $x^4$ is ${}^n C_4$.
• Coefficient of $x^5$ is ${}^n C_5$.
• Coefficient of $x^6$ is ${}^n C_6$.

Step 2: Apply the Arithmetic Progression Condition

The problem states that these coefficients are in an arithmetic progression (A.P.). For three numbers $a, b, c$ to be in A.P., the middle term $b$ must be the average of the other two terms, i.e., $2b = a+c$. Applying this to our coefficients: $2 \cdot {}^n C_5 = {}^n C_4 + {}^n C_6$

Step 3: Simplify the Equation using Properties of Binomial Coefficients

To solve this equation for $n$, we can use the property of binomial coefficients that relates consecutive terms: $\frac{{}^n C_r}{{}^n C_{r-1}} = \frac{n-r+1}{r}$ Let's rearrange the A.P. condition: $2 = \frac{{}^n C_4}{{}^n C_5} + \frac{{}^n C_6}{{}^n C_5}$ Now, we calculate the ratios:

• For the first ratio, we have ${}^n C_4$ and ${}^n C_5$. Here, $r=5$ and $r-1=4$. So, $\frac{{}^n C_5}{{}^n C_4} = \frac{n-5+1}{5} = \frac{n-4}{5}$ Therefore, $\frac{{}^n C_4}{{}^n C_5} = \frac{5}{n-4}$
• For the second ratio, we have ${}^n C_6$ and ${}^n C_5$. Here, $r=6$ and $r-1=5$. So, $\frac{{}^n C_6}{{}^n C_5} = \frac{n-6+1}{6} = \frac{n-5}{6}$

Substitute these ratios back into the A.P. equation: $2 = \frac{5}{n-4} + \frac{n-5}{6}$

Step 4: Solve the Algebraic Equation for $n$

Now, we solve this rational equation for $n$. First, find a common denominator, which is $6(n-4)$: $2 = \frac{5 \cdot 6 + (n-5)(n-4)}{6(n-4)}$ $2 = \frac{30 + n^2 - 4n - 5n + 20}{6n - 24}$ $2 = \frac{n^2 - 9n + 50}{6n - 24}$ Multiply both sides by $(6n-24)$: $2(6n - 24) = n^2 - 9n + 50$ $12n - 48 = n^2 - 9n + 50$ Rearrange into a standard quadratic equation ($Ax^2 + Bx + C = 0$): $n^2 - 9n - 12n + 50 + 48 = 0$ $n^2 - 21n + 98 = 0$

Now, we solve this quadratic equation. We can use factorization or the quadratic formula. Let's try factorization: We need two numbers that multiply to 98 and add to -21. These numbers are -7 and -14. $(n - 7)(n - 14) = 0$ This gives two possible values for $n$: $n = 7$ or $n = 14$

Step 5: Validate the Values of $n$

For binomial coefficients ${}^n C_r$ to be defined, $n$ must be a non-negative integer and $n \ge r$. In our problem, the highest value of $r$ used is 6 (for ${}^n C_6$).

• If $n=7$, then ${}^7 C_4, {}^7 C_5, {}^7 C_6$ are all valid. Let's check if they are in A.P.: ${}^7 C_4 = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35$ ${}^7 C_5 = {}^7 C_{7-5} = {}^7 C_2 = \frac{7 \cdot 6}{2 \cdot 1} = 21$ ${}^7 C_6 = {}^7 C_{7-6} = {}^7 C_1 = 7$ Check A.P.: $2 \cdot 21 = 35 + 7 \implies 42 = 42$. This is true. So, $n=7$ is a valid solution.

• If $n=14$, then ${}^{14} C_4, {}^{14} C_5, {}^{14} C_6$ are all valid. Let's check if they are in A.P.: ${}^{14} C_4 = \frac{14 \cdot 13 \cdot 12 \cdot 11}{4 \cdot 3 \cdot 2 \cdot 1} = 1001$ ${}^{14} C_5 = \frac{14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 2002$ ${}^{14} C_6 = \frac{14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 3003$ Check A.P.: $2 \cdot 2002 = 1001 + 3003 \implies 4004 = 4004$. This is also true. So, $n=14$ is a valid solution.

Step 6: Determine the Maximum Value of $n$

The problem asks for the maximum value of $n$. Comparing the two valid solutions, $n=7$ and $n=14$, the maximum value is $14$.

Tips and Common Mistakes:

• Factorial Expansion vs. Ratio Property: While expanding factorials ($\frac{n!}{r!(n-r)!}$) directly is always an option, using the ratio property ($\frac{{}^n C_r}{{}^n C_{r-1}} = \frac{n-r+1}{r}$) often simplifies the algebra significantly, especially when dealing with consecutive binomial coefficients in an equation. This avoids cumbersome factorial manipulations.
• Checking Validity of $n$: Always ensure that the derived values of $n$ are consistent with the definition of binomial coefficients (i.e., $n$ must be an integer and greater than or equal to the largest $r$ value involved). In this case, $n \ge 6$. Both $n=7$ and $n=14$ satisfy this condition.
• Algebraic Errors: Be extremely careful when expanding and combining terms in the quadratic equation. A single sign error can lead to incorrect roots.

Summary:

We used the definition of binomial coefficients and the condition for an arithmetic progression to set up an equation involving ${}^n C_4, {}^n C_5$, and ${}^n C_6$. By utilizing the ratio property of binomial coefficients, we transformed the equation into a quadratic in $n$. Solving this quadratic yielded two valid integer solutions, $n=7$ and $n=14$. The maximum among these was $14$.