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Let (1 + 2x + x^2)^4 = a_0 + a_1 x + a_2 x^2 + ... + a_8 x^8.  What is the value of a_1 + a_3 + a_5 + a_7?

 Jul 18, 2022
 #1
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(x^2 + 2x + 1)^4   =

 

[ ( x + 1)^2 ] ^4   =

 

(x + 1)^8 =   1 x^8  +  8x^7  +  28x^6 + 56x^5 + 70x^4 + 56x^3 + 28x^2 + 8x +  1

 

a1 =  8

a3 = 56

a5 = 56

a7 = 8 

 

Sum =   64 * 2  =   128

 

cool cool cool

 Jul 18, 2022

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