AP EAMCET · Maths · Binomial Theorem
In the binomial expansion of \((a-b)^n, n \geq 5\), the sum of the \(5^{\text {th }}\) and \(6^{\text {th }}\) terms is zero. Then the value of \(\frac{a}{b}\) is
- A \(\frac{n-2}{3}\)
- B \(\frac{n-4}{5}\)
- C \(\frac{n-5}{7}\)
- D \(\frac{n-1}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{n-4}{5}\)
Step-by-step Solution
Detailed explanation
\(T_5 = \binom{n}{4} a^{n-4} (-b)^4 = \binom{n}{4} a^{n-4} b^4\) \(T_6 = \binom{n}{5} a^{n-5} (-b)^5 = -\binom{n}{5} a^{n-5} b^5\) \(T_5 + T_6 = 0 \implies \binom{n}{4} a^{n-4} b^4 - \binom{n}{5} a^{n-5} b^5 = 0\) \(\binom{n}{4} a^{n-4} b^4 = \binom{n}{5} a^{n-5} b^5\)…
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