The sum of the coefficients of all even degre | vedclass

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Question

(B) Let $f(x) = (x + \sqrt{x^3 - 1})^6 + (x - \sqrt{x^3 - 1})^6$.Using the binomial expansion $(a+b)^n + (a-b)^n = 2 \sum_{k=0, 2, 4, ...} \binom{n}{k

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) Let $f(x) = (x + \sqrt{x^3 - 1})^6 + (x - \sqrt{x^3 - 1})^6$.Using the binomial expansion $(a+b)^n + (a-b)^n = 2 \sum_{k=0, 2, 4, ...} \binom{n}{k} a^{n-k} b^k$,we have:$f(x) = 2 [ \binom{6}{0} x^6 + \binom{6}{2} x^4 (x^3 - 1) + \binom{6}{4} x^2 (x^3 - 1)^2 + \binom{6}{6} (x^3 - 1)^3 ]$$f(x) = 2 [ 1 \cdot x^6 + 15 x^4 (x^3 - 1) + 15 x^2 (x^6 - 2x^3 + 1) + 1 (x^9 - 3x^6 + 3x^3 - 1) ]$$f(x) = 2 [ x^6 + 15x^7 - 15x^4 + 15x^8 - 30x^5 + 15x^2 + x^9 - 3x^6 + 3x^3 - 1 ]$$f(x) = 2 [ x^9 + 15x^8 + 15x^7 - 2x^6 - 30x^5 - 15x^4 + 3x^3 + 15x^2 - 1 ]$Expanding the terms,the even degree terms are $15x^8, -2x^6, -15x^4, 15x^2, -1$ (where $-1$ is $x^0$).The coefficients are $15, -2, -15, 15, -1$.Sum of coefficients $= 2 	imes (15 - 2 - 15 + 15 - 1) = 2 	imes 12 = 24$.

The sum of the coefficients of all even degree terms in $x$ in the expansion of $(x + \sqrt{x^3 - 1})^6 + (x - \sqrt{x^3 - 1})^6$ for $x > 1$ is equal to:

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