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7.Binomial Theorem
hard
અહી ${ }^{n} C_{r}$ એ $(1+ x )^{ n }$ ના વિસ્તરણમાં $x^{r}$ નો સહગુણક દર્શાવે છે. જો $\sum_{ k =0}^{10}\left(2^{2}+3 k \right){ }^{ n } C _{ k }=\alpha .3^{10}+\beta \cdot 2^{10}, \alpha, \beta \in R$ તો $\alpha+\beta$ ની કિમંત મેળવો.
A
$19$
B
$21$
C
$17$
D
$13$
(JEE MAIN-2021)
Solution
Instead of ${ }^{n} C_{k}$ it must be ${ }^{10} C_{k}$ i.e.
$\sum_{k=0}^{10}\left(2^{2}+3 k\right){ }^{10} C _{ k }=\alpha .3^{10}+\beta .2^{10}$
$LHS =4 \sum_{ k =0}^{10}{ }^{10} C _{ k }+3 \sum_{ k =0}^{10} k \cdot \frac{10}{ k } \cdot{ }^{9} C _{ k -1}$
$=4.2^{10}+3.10 .2^{9}$
$=19.2^{10}=\alpha .3^{10}+\beta .2^{10}$
$\Rightarrow \alpha=0, \beta=19 \Rightarrow \alpha+\beta=19$
Standard 11
Mathematics
