Given the polynomial (1+x)^6 calculate the sum of even coefficients of polynomial.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll expand the binomial and we'll get:

(1+x)^6 = a0 + a1*x + a2*x^2 + ... + a6*x^6

a0,a1,a2,...,a6 are the coefficients of polynomial.

The sum of even coefficients is:

a0 + a2 + a4 + a6

We know that if we want to determine the summof all coefficients of a polynomial, we'll have to make the variable x = 1.

We'll put x = 1 and we'll calculate:

(1 + 1)^6 = a0 + a1*1 + a2*1^2 + ... + a6*1^6

a0  +a1 + ... + a6 = 2^6 (1)

Now, we'll put x = -1

(1 - 1)^6 = a0  -a1 + ... - a5 + a6

a0  -a1 + ... - a5 + a6 = 0 (2)

We'll add (1) + (2):

a0  +a1 + ... + a6 + a0  -a1 + ... - a5 + a6 = 2^6

We'll eliminate and combine like terms:

2(a0 + a2 + a4 + a6) = 2^6

a0 + a2 + a4 + a6 = 2^6/2

a0 + a2 + a4 + a6 = 2^5

The sum of even coefficientas of the given polynomial is: a0 + a2 + a4 + a6 = 2^5

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