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Sum of Coefficients of All Terms

If C0, C2, ...

Question

If $C_{0}, C_{2}, C_{4}, . . . . . . . . . .$ are binomial coefficients in the expansion of $(1 + x)^{9}$ , then $C_{0} + C_{2} + C_{4} + C_{6} + C_{8} =$

A

2^{7}

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B

256

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C

2^{9}

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D

258

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Solution

The correct option is B $256$
$(1 + x)^{9} =$ $= 1 +^{9} C_{1} x +^{9} C_{2} x^{2} +^{9} C_{3} x^{3} +^{9} C_{4} x^{4} +^{9} C_{5} x^{5} +^{9} C_{6} x^{6} +^{9} C_{7} x^{7} +^{9} C_{8} x^{8} +^{9} C_{9} x^{9}$
Putting $x = 1$ , we get
$2^{9} = 1 +^{9} C_{1} +^{9} C_{2} +^{9} C_{3} +^{9} C_{4} +^{9} C_{5} +^{9} C_{6} +^{9} C_{7} x^{7} +^{9} C_{8} +^{9} C_{9}$ ...(i)
Putting $x = - 1$ , we get
$0 = 1 -^{9} C_{1} +^{9} C_{2} -^{9} C_{3} +^{9} C_{4} -^{9} C_{5} +^{9} C_{6} -^{9} C_{7} x^{7} +^{9} C_{8} -^{9} C_{9}$ ...(ii)
Adding (i) and (ii), we get
$2^{9} = 2 [1 +^{9} C_{2} +^{9} C_{4} +^{9} C_{6} +^{9} C_{8}]$
$2^{8} =^{9} C_{0} +^{9} C_{2} +^{9} C_{4} +^{9} C_{6} +^{9} C_{8}$
Therefore,
$^{9} C_{0} +^{9} C_{2} +^{9} C_{4} +^{9} C_{6} +^{9} C_{8} = 2^{8}$ $= 256$

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