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7.Binomial Theorem
hard
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
$12.75$
(JEE MAIN-2019)
Solution
$(10+x)^{50}+(10-x)^{50}$
$a_{0}=\left(10^{50}\right)(2)$
$a_{2}=^{50} C_{2}(10)^{48}(2)$
$\frac{a_{2}}{a_{0}}=\frac{^{50} C_{2}(10)^{48}(2)}{10^{52}(2)}=12.25$
Standard 11
Mathematics
