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7.Binomial Theorem
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Let $\left(\frac{n}{k}\right)=\frac{n !}{k !(n-k) !}$. Then the sum $\frac{1}{2^{10}} \sum \limits_{ k =0}^{10}\left(\frac{10}{ k }\right) k ^2$, lies in the interval
A
$(26,27)$
B
$(27,28)$
C
$(28,29)$
D
$(29,30)$
(KVPY-2021)
Solution
(b)
$\sum \limits_{ k =0}^{10}{ }^{10} C _{ k } \cdot k ^2=\sum \limits_{ k =1}^{10} \frac{10}{ k } \cdot{ }^9 C _{ k -1} \cdot k ^2$
$10 \sum\limits_{ k =1}^{10}( k -1+1)^9 C _{ k -1}=10\left(\sum \limits_{ k =1}^{10} 9 \cdot{ }^8 C _{ k -2}+\sum \limits_{ k =1}^{10}{ }^9 C _{ k -1}\right)$
$10\left(9 \cdot 2^8+2^9\right)=10 \cdot 2^8(9+2)=110 \cdot 2^8$
$\frac{1}{2^{10}} \times 28 \times 110=\frac{2^9}{2^{10}} \cdot(55)=\frac{55}{2}$
Standard 11
Mathematics
