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7.Binomial Theorem
hard
$n$ ની ન્યૂનતમ કિમંત મેળવો કે જેથી દ્રીપદી વિસ્તરણમાં $(\sqrt[3]{7}+\sqrt[12]{11})^{ n }$ માં પૃણાંક પદોની સંખ્યા $183$ મળે.
A$2184$
B$2148$
C$2172$
D$2196$
(JEE MAIN-2025)
Solution
General term $={ }^n C _{ r }\left\{7^{1 / 3}\right\}^{ n – r }\left(11^{1 / 12}\right)^{ r }$
$={ }^{n} C_{r}\{7\}^{\frac{n-r}{3}}(11)^{r / 12}$
For integral terms, r must be multiple of $12$
$\therefore r=12 k, k \in W$
Total values of $r =183$
Hence $\max r =12(182)$
$=2184$
Min value of $n=2184$
$={ }^{n} C_{r}\{7\}^{\frac{n-r}{3}}(11)^{r / 12}$
For integral terms, r must be multiple of $12$
$\therefore r=12 k, k \in W$
Total values of $r =183$
Hence $\max r =12(182)$
$=2184$
Min value of $n=2184$
Standard 11
Mathematics
