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Uchiha Sasuke

25/05/2018 at 08:39
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Let P(x) be the polynomial given by \(P\left(x\right)=\left(2+x+2x^3\right)^{15}\). Suppose that \(P\left(x\right)=a_0+a_1x+a_2x^2+...+a_{45}x^{45}\).The value of \(S=a_1-a_2+a_3-a_4+...-a_{44}+a_{45}\).


Polynomial

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    Uchiha Sasuke 26/05/2018 at 01:19

    Thank you very much yeu

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    Dao Trong Luan Coordinator 25/05/2018 at 12:30

    P(x) = a0 + a1x + a2x2 + ... + a45x45

    => P(1) = a0 + a1 + a2 + ... + a45 = (2 + 1 + 2)15 = 515

    => P(-1) = a0 - a1 + a2 - ... - a45 = (2 - 1 - 2)15 = (-1)15

    ==> P(1) - P(-1) = 2a1 + 2a3 + ... + 2a45 = 515 + 1

            P(1) + P(-1) = 2a0 + 2a2 + ... + 2a44 = 515 - 1

    => (a1 + a3 + ... + a45) - (a0 + a2 + ... + a44) = \(\dfrac{5^{15}+1-5^{15}+1}{2}=1\)

    => S = -a0 + a1 - a2 + .... - a44 + a45 = 1

    P/s: Not S in the topic banh



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