Key Concepts and Formulas

The Binomial Theorem states that for any positive integer $n$, the expansion of $(x+y)^n$ is given by: $(x+y)^n = \sum_{r=0}^{n} {}^nC_r x^{n-r} y^r = {}^nC_0 x^n y^0 + {}^nC_1 x^{n-1} y^1 + \dots + {}^nC_n x^0 y^n$ The coefficients in this expansion are ${}^nC_0, {}^nC_1, \dots, {}^nC_n$.

1. Sum of Coefficients: To find the sum of the coefficients in the expansion of $(x+y)^n$, we substitute $x=1$ and $y=1$ into the expansion. Sum of coefficients $= (1+1)^n = 2^n$.

2. Greatest Coefficient: In the expansion of $(x+y)^n$:

• If $n$ is an even integer, the greatest coefficient is the coefficient of the middle term, which is ${}^nC_{n/2}$.
• If $n$ is an odd integer, there are two middle terms, and their coefficients are equal and greatest: ${}^nC_{(n-1)/2}$ and ${}^nC_{(n+1)/2}$.

Step-by-Step Solution

Step 1: Determine the value of 'n'

We are given that the sum of the coefficients in the expansion of $(x+y)^n$ is $4096$. Why this step? To find the greatest coefficient, we first need to know the value of $n$.

Using the formula for the sum of coefficients: $2^n = 4096$

Now, we need to express $4096$ as a power of $2$:

• $2^1 = 2$
• $2^2 = 4$
• ...
• $2^{10} = 1024$
• $2^{11} = 2048$
• $2^{12} = 4096$

So, we have: $2^n = 2^{12}$ Equating the exponents, we get: $n = 12$

Step 2: Identify the greatest coefficient for the determined 'n'

Since $n=12$ (an even number), the greatest coefficient in the expansion of $(x+y)^{12}$ will be the coefficient of the middle term. Why this step? The greatest coefficient formula depends on whether $n$ is even or odd. Since $n=12$ is even, we use the corresponding formula.

For an even $n$, the greatest coefficient is ${}^nC_{n/2}$. Substituting $n=12$: Greatest coefficient $= {}^{12}C_{12/2} = {}^{12}C_6$.

Step 3: Calculate the value of the greatest coefficient

Now we calculate the value of ${}^{12}C_6$. Why this step? This is the final calculation to arrive at the answer.

The formula for combinations is ${}^nC_r = \frac{n!}{r!(n-r)!}$. ${}^{12}C_6 = \frac{12!}{6!(12-6)!} = \frac{12!}{6!6!}$ Expanding the factorials: ${}^{12}C_6 = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6!}{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 6!}$ Cancel out $6!$ from the numerator and denominator: ${}^{12}C_6 = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$ Simplify the expression: ${}^{12}C_6 = \frac{\cancel{12}}{\cancel{6 \times 2}} \times 11 \times \frac{\cancel{10}}{\cancel{5}} \times \frac{\cancel{9}}{\cancel{3}} \times \frac{\cancel{8}}{\cancel{4}} \times 7$ ${}^{12}C_6 = 1 \times 11 \times 2 \times 3 \times 2 \times 7$ ${}^{12}C_6 = 924$

Tips and Common Mistakes

• Tip for powers of 2: It's very helpful to memorize common powers of 2 (e.g., $2^{10} = 1024$) to quickly solve equations involving powers of 2.
• Common Mistake (Greatest Coefficient vs. Greatest Term): Do not confuse the "greatest coefficient" with the "greatest term". The greatest coefficient is purely based on $n$, while the greatest term in the expansion depends on the actual values of $x$ and $y$. For example, in $(2+3)^2 = 4+12+9$, the greatest coefficient is ${}^2C_1=2$, but the greatest term is $12$. Here, the question specifically asks for the "greatest coefficient".
• Combinations Calculation: Be careful with factorials and cancellations to avoid arithmetic errors.

Summary/Key Takeaway

This problem effectively tests your understanding of two fundamental properties of binomial expansions: how to calculate the sum of coefficients by setting variables to 1, and how to identify the greatest coefficient based on whether $n$ is even or odd. Once $n$ is determined, the problem reduces to a straightforward calculation of a combination ${}^nC_r$.