Let left(frac{n}{k}right)=frac{n !}{k !(n-k) !} . Then sum frac{1}{2^{10}} Σ limits

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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

$(26,27)$

$(27,28)$

$(28,29)$

$(29,30)$

(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

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(AIEEE-2012)

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hard

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medium

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(AIEEE-2011)

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

Solution

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$\left( {\begin{array}{{20}{c}}n\\0\end{array}} \right) + 2\,\left( {\begin{array}{{20}{c}}n\\1\end{array}} \right) + {2^2}\left( {\begin{array}{{20}{c}}n\\2\end{array}} \right) + ….. + {2^n}\left( {\begin{array}{{20}{c}}n\\n\end{array}} \right)$ is equal to