JEE Mains · Maths · STD 11 - 7. binomial theoram
If the number of terms in the expansion of \({\left( {1 - \frac{2}{x} + \frac{4}{{{x^2}}}} \right)^n},x \ne 0\) is \(28\) then the sum of the coefficients of all the terms in this expansion, is :
- A \(243\)
- B \(729\)
- C \(64\)
- D \(2187\)
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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