JEE Mains · Maths · STD 11 - 7. binomial theoram
Let \({\left( {x + 10} \right)^{50}} + {\left( {x - 10} \right)^{50}} = {a_0} + {a_1}x + {a_2}{x^2} + .... + {a_{50}}{x^{50}}\) , for \(x \in R\); then \(\frac{{{a_2}}}{{{a_0}}}\) is equal to
- A \(12.50\)
- B \(12\)
- C \(12.25\)
- D \(12.75\)
Answer & Solution
Correct Answer
(C) \(12.25\)
Step-by-step Solution
Detailed explanation
\((10+x)^{50}+(10-x)^{50}\) \(a_{0}=\left(10^{50}\right)(2)\) \(a_{2}=^{50} C_{2}(10)^{48}(2)\) \(\frac{a_{2}}{a_{0}}=\frac{^{50} C_{2}(10)^{48}(2)}{10^{52}(2)}=12.25\)
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