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7.Binomial Theorem
hard
If $\sum_{ k =1}^{10} K ^{2}\left(10_{ C _{ K }}\right)^{2}=22000 L$, then $L$ is equal to $.....$
A
$222$
B
$221$
C
$223$
D
$224$
(JEE MAIN-2022)
Solution
$\sum_{ K =1}^{10} K ^{2}\left({ }^{10} C _{ K }\right)^{2}$
$\sum_{ K =1}^{10}\left( K ^{10} C _{ K }\right)^{2}=\sum_{ K =1}^{10}\left(10 \cdot{ }^{9} C _{ K -1}\right)^{2}$
$=100 \sum_{ K =1}^{9} C _{ K -1} \cdot{ }^{9} C _{10- K }$
$=100\left({ }^{18} C _{9}\right)=100\left(\frac{18 !}{9 ! 9 !}\right)$
$\Rightarrow 4862000=22000 L$
Hence $L =221$
Standard 11
Mathematics
