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7.Binomial Theorem
hard
$\left(3^{\frac{1}{2}}+5^{\frac{1}{4}}\right)^{680}$ ના વિસ્તરણમાં પૂર્ણાક પદોની સંખ્યા $..........$ છે.
A
$170$
B
$171$
C
$172$
D
$173$
(JEE MAIN-2023)
Solution
The number of integral term in the expression of $\left(3^{\frac{1}{2}}+5^{\frac{1}{4}}\right)^{680}$ is equal to
General term $={ }^{680} C _{ r }\left(3^{\frac{1}{2}}\right)^{680- r }\left(5^{\frac{1}{4}}\right)^{ r }$
$={ }^{680} C _{ r } 3^{\frac{680- r }{2}} 5^{\frac{ r }{4}}$
Value's of $r$, where $\frac{r}{4}$ goes to integer
$r =0,4,8,12, \ldots \ldots \ldots .680$
All value of $r$ are accepted for $\frac{680-r}{2}$ as well so No of integral terms $=171$.
Standard 11
Mathematics
