ExamPlay Dark Logo
登入

JEE Advance - Mathematics (2003 - No. 19)

Coefficient of $${t^{24}}$$ in $${\left( {1 + {t^2}} \right)^{12}}\left( {1 + {t^{12}}} \right)\left( {1 + {t^{24}}} \right)$$ is
$${}^{12}{C_6} + 3$$
$${}^{12}{C_6} + 1$$
$${}^{12}{C_6}$$
$${}^{12}{C_6} + 2$$

评论 (0)

登录发表评论
广告
BrainBehindX Inc Logo
©2026; 供电 BrainBehindX Inc