If 1-x-2 x26-1-a1 x-a2 x2-a3 x3-a12x12- then the value ofa2-a4-a6-a12 will be

(1+x 2 x^2)^6=1+a_1 x+a_2 x^2+⋯+a_12x^12 Putting x=1, we get 0=1+a_1+a_2+a_3+⋯+a_12⋯(1) Putting x= 1, we get 64=1 a_1+a_2 a_3+⋯+a_12 (1)+(2) gives 64=2[1+a_2+a_ ...

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Byju's Answer

Standard XIII

Mathematics

Sum of Binomial Coefficients of Even Numbered Terms

If 1+x-2 x26=...

Question

If ${(1 + x - 2 x^{2})}^{6} = 1 + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots + a_{12} x^{12},$ then the value of
$a_{2} + a_{4} + a_{6} + \dots + a_{12}$ will be

32

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31

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64

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1024

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Solution

The correct option is B $31$
${(1 + x - 2 x^{2})}^{6} = 1 + a_{1} x + a_{2} x^{2} + \dots + a_{12} x^{12}$
Putting $x = 1,$ we get
$0 = 1 + a_{1} + a_{2} + a_{3} + \dots + a_{12} \dots (1)$
Putting $x = - 1,$ we get $64 = 1 - a_{1} + a_{2} - a_{3} + \dots + a_{12}$
$(1) + (2)$ gives
$64 = 2 [1 + a_{2} + a_{4} + \dots a_{12}] \dots (2)$
$\Rightarrow 1 + a_{2} + a_{4} + \dots + a_{12} = 32$
$\Rightarrow a_{2} + a_{4} + \dots + a_{32} = 31$

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If (1+x−2x2)6=1+a1x+a2x2+a3x3+⋯+a12x12, then the value of a2+a4+a6+⋯+a12 will be

The correct option is B 31(1+x−2x2)6=1+a1x+a2x2+⋯+a12x12 Putting x=1, we get 0=1+a1+a2+a3+⋯+a12⋯(1) Putting x=−1, we get 64=1−a1+a2−a3+⋯+a12 (1)+(2) gives 64=2[1+a2+a4+⋯a12]⋯(2) ⇒1+a2+a4+⋯+a12=32 ⇒a2+a4+⋯+a32=31

If ${(1 + x - 2 x^{2})}^{6} = 1 + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots + a_{12} x^{12},$ then the value of
$a_{2} + a_{4} + a_{6} + \dots + a_{12}$ will be