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JEE Mains · Maths · STD 11 - 7. binomial theoram

The absolute difference of the coefficients of \(x^{10}\) and \(x^7\) in the expansion of \(\left(2 x^2+\frac{1}{2 x}\right)^{11}\) is equal to

  1. A \(12^3-12\)
  2. B \(11^3-11\)
  3. C \(10^3-10\)
  4. D \(13^3-13\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(12^3-12\)

Step-by-step Solution

Detailed explanation

\(T _{ r +1}={ }^{11} C _{ r }\left(2 x ^2\right)^{11- r }\left(\frac{1}{2 x }\right)^{ T }\) \(={ }^{11} C _{ r } 2^{11-2 r } x ^{22-3 r }\) \(22-3 r =10 \quad \text { and } \quad 22-3 r =7\) \(r =4 \quad \text { and } \quad r =5\)…
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