If ${C_{r}}^{13}$ denoted by $C_{r}$ then 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 C $2^{12} - 287$
Consider
$(1 + x)^{13} =^{13} C_{0} +^{13} C_{1} . x^{1} +^{13} C_{2} . x^{2} +^{13} C_{3} . x^{3} . . . +^{13} C_{13} . x^{13}$
Now Let x=1.
$2^{13} =^{13} C_{0} +^{13} C_{1} +^{13} C_{2} +^{13} C_{3} . . . +^{13} C_{13}$
$= 2 [^{13} C_{0} +^{13} C_{1} +^{13} C_{2} +^{13} C_{3} . . . +^{13} C_{6}]$
Hence
$2^{12} = a_{0} + a_{1} + . . . a_{6}$ ...(i)
Similarly
$(1 - x)^{13} |_{x = - 1} = a_{0} - a_{1} + a_{2} - a_{3} . . . a_{6} = 0... (i i)$
Adding i and ii, we get
$2^{12} = 2 (a_{0} + a_{2} + a_{4} + a_{6})$
$= 2 (a_{0} + a_{11} + a_{9} + a_{7})$
Hence
$2^{11} = a_{0} + a_{11} + a_{9} + a_{7}$ ...(a)
Subtracting ii from i, we get
$2^{12} = 2 (a_{1} + a_{3} + a_{5})$
Or
$2^{11} = a_{1} + a_{3} + a_{5}$ ...(b)
Adding a and b, we get
$2^{12} = a_{1} + a_{5} + a_{7} + a_{9} + (a_{0} + a_{3})$
$2^{12} = a_{1} + a_{5} + a_{7} + a_{9} + (1 +^{13} C_{3})$
$2^{12} = a_{1} + a_{5} + a_{7} + a_{9} + (287)$
$2^{12} - 287 = a_{1} + a_{5} + a_{7} + a_{9}$ .

Suggest Corrections

Similar questions

^{13} C_{r}

C_{r}

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}

c_{0}, c_{1}, c_{2} \dots c_{n}

{(1 + x)}^{n}

c_{1} + c_{5} + c_{9} + c_{13} + \dots

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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