Question

Numerically greatest term in the expansion of $(2x-3y)^n$ when $x=\frac{7}{5}, y=\frac{3}{7}$ and $n=13$ is

• A. $13.3^5.7^9$
• B. $13.3^4.7^9$
• C. $26.3^5.7^9$
• D. $26.3^4.7^9$

Hint:

When calculating the numerically greatest term, be very careful with the formula for the ratio, which is $\frac{n-r+1}{r} |\frac{B}{A}|$. A common mistake is using $r+1$ in the denominator. If your calculated value does not match the options, double-check your arithmetic. If it still doesn't match, the question might be flawed, which can happen in competitive exams.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
To find the numerically greatest term in the binomial expansion of $(a+b)^n$, we find the value of $r$ for which the term $T_{r+1} = {^n}C_r a^{n-r} b^r$ is maximum. This is done by analyzing the ratio of consecutive terms, $|\frac{T_{r+1}}{T_r}|$.
Step 2: Key Formula or Approach
The numerically greatest term is $T_{r+1}$ where $r$ is determined by the inequality:
\[ \left|\frac{T_{r+1}}{T_r}\right| \ge 1 \] For the expansion $(A+B)^n$, this ratio is $\frac{n-r+1}{r}\left|\frac{B}{A}\right|$. We solve for $r$ using this inequality.
If the result is $r \le m$, where $m$ is not an integer, the greatest term is $T_{\lfloor m \rfloor+1}$. If $m$ is an integer, $T_m$ and $T_{m+1}$ are equal and are the greatest terms.
Step 3: Detailed Explanation
The expansion is for $(2x-3y)^{13}$.
Here, $n=13$, $A = 2x$, and $B = -3y$.
Given $x = \frac{7}{5}$ and $y = \frac{3}{7}$.
First, calculate the values of A and B:
\[ A = 2x = 2 \cdot \frac{7}{5} = \frac{14}{5} \] \[ B = -3y = -3 \cdot \frac{3}{7} = -\frac{9}{7} \] Now, we set up the inequality:
\[ \frac{13-r+1}{r} \left| \frac{-9/7}{14/5} \right| \ge 1 \] \[ \frac{14-r}{r} \left( \frac{9/7}{14/5} \right) \ge 1 \] \[ \frac{14-r}{r} \left( \frac{9 \times 5}{7 \times 14} \right) \ge 1 \] \[ \frac{14-r}{r} \left( \frac{45}{98} \right) \ge 1 \] \[ 45(14-r) \ge 98r \] \[ 630 - 45r \ge 98r \] \[ 630 \ge 143r \] \[ r \le \frac{630}{143} \approx 4.405 \] Since $r$ must be an integer, the largest value of $r$ satisfying this is $r=4$.
The numerically greatest term is $T_{r+1} = T_{4+1} = T_5$.
Now we calculate the value of $T_5$:
\[ T_5 = {^{13}}C_4 (2x)^{13-4} (-3y)^4 = {^{13}}C_4 (2x)^9 (3y)^4 \] \[ {^{13}}C_4 = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 13 \times 5 \times 11 = 715 \] \[ T_5 = 715 \left(2 \cdot \frac{7}{5}\right)^9 \left(3 \cdot \frac{3}{7}\right)^4 = 715 \left(\frac{14}{5}\right)^9 \left(\frac{9}{7}\right)^4 \] \[ T_5 = 715 \frac{14^9}{5^9} \frac{9^4}{7^4} = (5 \times 11 \times 13) \frac{(2 \times 7)^9}{5^9} \frac{(3^2)^4}{7^4} \] \[ T_5 = (5 \times 11 \times 13) \frac{2^9 \times 7^9}{5^9} \frac{3^8}{7^4} = \frac{11 \times 13 \times 2^9 \times 3^8 \times 7^5}{5^8} \] This calculated value does not match any of the options. This indicates a probable typo in the question's parameters ($x, y, n$) or the options provided.
Given the discrepancy, it's impossible to logically derive the provided correct answer from the question as stated. There must be an error in the source material.
Step 4: Final Answer
Based on a correct application of the standard method, the numerically greatest term is the 5th term, $T_5$, whose value is $\frac{11 \times 13 \times 2^9 \times 3^8 \times 7^5}{5^8}$. This does not match the provided options, suggesting an error in the problem statement.