JEE Mains · Maths · STD 11 - 7. binomial theoram
Let \(\left(2 x ^{2}+3 x +4\right)^{10}=\sum \limits_{ r =0}^{20} a _{ r } x ^{ r } \cdot\) Then \(\frac{ a _{7}}{ a _{13}}\) is equal to
- A \(4\)
- B \(32\)
- C \(16\)
- D \(8\)
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
Given \(\left(2 x^{2}+3 x+4\right)^{10}=\sum_{ r =0}^{20} a _{ r } x ^{ r }\) replace x by \(\frac{2}{x}\) in above identity :- \(\frac{2^{10}\left(2 x ^{2}+3 x +4\right)^{10}}{ x ^{20}}=\sum_{ r =0}^{20} \frac{ a _{ r } 2^{ r }}{ x ^{ r }}\)…
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