If $^{13} C_{r}$ is denoted by $C_{r}$ , then the value of $C_{1} + C_{5} + C_{7} + C_{9} + C_{11}$ is equal to

2^{12} - 287

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2^{12} - 165

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2^{12} - C_{3}

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2^{12} - C_{2} - C_{13}

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Solution

The correct option is A $2^{12} - 287$
$C_{1} + C_{3} + C_{5} + C_{7} + C_{9} + C_{11} + C_{13} = C_{0} + C_{2} + C_{4} + C_{6} + C_{8} + C_{10} + C_{12} = 2^{12}$
$∴$ $C_{1} + C_{5} + C_{7} + C_{9} + C_{11} = 2^{12} - C_{3} - C_{13}$
$= 2^{12} - \frac{13 \times 12 \times 11}{6} - 1$
$= 2^{12} - 286 - 1 = 2^{12} - 287$

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Similar questions

{C_{r}}^{13}

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c_{1} + c_{5} + c_{7} + c_{9} + c_{11}

z

(1 + z^{2} + z^{4})^{8} = C_{0} + C_{1} z^{2} + C_{2} z^{4} + . . . + C_{16} z^{32}

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212×424+212×366+212×240

Explain With Solution

{(\frac{- 1 + i \sqrt{3}}{2})}^{12} + {(\frac{- 1 - i \sqrt{3}}{2})}^{12}

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