COMEDK · Maths · 7. Binomial Theorem
Middle term in the expansion of \(\left(x^{2}+\dfrac{1}{x^{2}}+2\right)^{n}\) is
- A \(\dfrac{n !}{(n !)^{2}}\)
- B \(\dfrac{(2 n) !}{(n !)^{2}}\)
- C \(\dfrac{(2 n-1) !}{n !}\)
- D \(\dfrac{(2 n) !}{n !}\)
Answer & Solution
Correct Answer
(B) \(\dfrac{(2 n) !}{(n !)^{2}}\)
Step-by-step Solution
Detailed explanation
The given expression is \(\left(x^{2} + \dfrac{1}{x^{2}} + 2\right)^{n}\). This can be rewritten as \(\left(x + \dfrac{1}{x}\right)^{2n}\). The number of terms in the expansion of \(\left(x + \dfrac{1}{x}\right)^{2n}\) is \(2n + 1\), which is odd. Therefore, there is only one…
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