Question:
The constant term in the binomial expansion of \(\left(2x - \frac{5}{x^2}\right)^6\) is
The constant term in the binomial expansion of \(\left(2x - \frac{5}{x^2}\right)^6\) is
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For \((ax^p + bx^q)^n\), constant term occurs when \(p(n-r) + qr = 0\).
Updated On: Apr 25, 2026
- 4800
- 3200
- 5600
- 5400
- 6000
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The Correct Option is
Solution and Explanation
Step 1: Understanding the Concept:
General term: \(T_{r+1} = \binom{6}{r} (2x)^{6-r} \left(-\frac{5}{x^2}\right)^r = \binom{6}{r} 2^{6-r} (-5)^r x^{6-r-2r}\).
Step 2: Detailed Explanation:
Power of x: \(6 - 3r = 0 \implies r=2\). So constant term = \(\binom{6}{2} 2^{4} (-5)^2 = 15 \times 16 \times 25 = 6000\).
Step 3: Final Answer:
Option (E).
General term: \(T_{r+1} = \binom{6}{r} (2x)^{6-r} \left(-\frac{5}{x^2}\right)^r = \binom{6}{r} 2^{6-r} (-5)^r x^{6-r-2r}\).
Step 2: Detailed Explanation:
Power of x: \(6 - 3r = 0 \implies r=2\). So constant term = \(\binom{6}{2} 2^{4} (-5)^2 = 15 \times 16 \times 25 = 6000\).
Step 3: Final Answer:
Option (E).
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