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

यदि \(\left(1-\frac{2}{x}+\frac{4}{x^{2}}\right)^{n}, x \neq 0\) के प्रसार में पदों की संख्या \(28\) है, तो इस प्रसार में आने वाले सभी पदों के गुणांकों का योग है:

  1. A \(243\)
  2. B \(729\)
  3. C \(64\)
  4. D \(2187\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(729\)

Step-by-step Solution

Detailed explanation

Clearly, number of terms in the expansion of \(\left(1-\frac{2}{x}+\frac{4}{x^{2}}\right)^{n}\) is \(\frac{(n+2)(n+1)}{2}\) or \(^{n+2} C_{2}\) [assuming \(\left.\frac{1}{x} \text { and } \frac{1}{x^{2}} \text { distinct }\right]\) \(\therefore \frac{(n+2)(n+1)}{2}=28\)…
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