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_ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

31

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

64

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1024

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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
$64 = 2 [1 + a_{2} + a_{4} + \dots a_{12}] \dots (2)$
$\Rightarrow 1 + a_{2} + a_{4} + \dots + a_{12} = 32$
$\Rightarrow a_{2} + a_{4} + \dots + a_{32} = 31$

Suggest Corrections

Similar questions

Q. If

{(1 + x - 2 x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}

and the value of

a_{1} + a_{3} + a_{5} + \dots + a_{39} = - 2^{k}

, then

k =

Q. If

(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . . + a_{12} x^{12}

, then

a_{2} + a_{4} + a_{6} + . . . . + a_{12} =

Q. If

(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . . . . + a_{12} x^{12}

, then the expression

a_{2} + a_{4} + a_{6} + . . . . . . + a_{12}

has the value

Q. If

(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + . . . . + a_{12} x^{12}

, then the expression

a_{2} + a_{4} + a_{6} + . . . + a_{12}

has the value.

Q. Let

A_{1} = [a_{1}] A_{2} = [\begin{matrix} a_{2} & a_{3} a_{4} & a_{5} \end{matrix}] A_{3} = ⎡ ⎢ ⎣ \begin{matrix} a_{6} & a_{7} & a_{8} a_{9} & a_{10} & a_{11} a_{12} & a_{13} & a_{14} \end{matrix} ⎤ ⎥ ⎦ \dots \dots A_{n} = [\dots] where a_{r} = [{log}_{2} r]

([\cdot]

denotes the greatest integer

)

. Then trace of

A_{10}

Join BYJU'S Learning Program

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