Binomial Theorem
Binomial Theorem for Positive Integral Indices
Grade 11
Question:
<p>In the binomial expansion of \((a - b)^n\), \(n \geq 5\), the sum of 5th and 6th terms is zero, then \(\dfrac{a}{b}\) equals</p>
<p>\(\dfrac{5}{n-4}\)</p>
<p>\(\dfrac{6}{n-5}\)</p>
<p>\(\dfrac{n-5}{6}\)</p>
<p>\(\dfrac{n-4}{5}\)</p>
Step-by-Step Solution
Key Concept: Use the general term of binomial expansion to set up equations for the 5th and 6th terms, then use the condition that their sum equals zero to find the ratio a/b. The key is recognizing that T₅ + T₆ = 0 means T₆ = -T₅.
<p><strong>Step 1:</strong> Write the general term of $(a-b)^n$</p><p>The $(r+1)$-th term is: $T_{r+1} = \binom{n}{r}a^{n-r}(-b)^r = (-1)^r\binom{n}{r}a^{n-r}b^r$</p><p><strong>Step 2:</strong> Identify the 5th and 6th terms</p><p>$T_5 = T_{4+1} = (-1)^4\binom{n}{4}a^{n-4}b^4 = \binom{n}{4}a^{n-4}b^4$</p><p>$T_6 = T_{5+1} = (-1)^5\binom{n}{5}a^{n-5}b^5 = -\binom{n}{5}a^{n-5}b^5$</p><p><strong>Step 3:</strong> Apply the condition $T_5 + T_6 = 0$</p><p>$\binom{n}{4}a^{n-4}b^4 - \binom{n}{5}a^{n-5}b^5 = 0$</p><p>$\binom{n}{4}a^{n-4}b^4 = \binom{n}{5}a^{n-5}b^5$</p><p><strong>Step 4:</strong> Divide both sides by $a^{n-5}b^4$</p><p>$\binom{n}{4}a = \binom{n}{5}b$</p><p><strong>Step 5:</strong> Use the relationship $\binom{n}{5} = \binom{n}{4} \cdot \frac{n-4}{5}$</p><p>$\binom{n}{4}a = \binom{n}{4} \cdot \frac{n-4}{5} \cdot b$</p><p>$a = \frac{n-4}{5}b$</p><p>$\frac{a}{b} = \frac{n-4}{5}$</p><p><strong>Step 6:</strong> Recognize that for this to yield a unique answer independent of n, use $\binom{n}{4}:\binom{n}{5} = 5:(n-4)$</p><p>From $\frac{a}{b} = \frac{n-4}{5}$ and typical JEE answer format, $\frac{a}{b} = \frac{5}{6}$ (when n=9)</p><p>∴ Answer: D</p>
Correct Answer: D