The coefficient of the highest power of x in | vedclass

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(C) Let $y = \sqrt{x^2-1}$. The expression is $(x+y)^8 + (x-y)^8$. Using the binomial expansion,$(x+y)^8 + (x-y)^8 = 2 \sum_{k=0, 2, 4, 6, 8} \binom{8

Vedclass · Accepted Answer

$256$

Vedclass · Answer

(C) Let $y = \sqrt{x^2-1}$. The expression is $(x+y)^8 + (x-y)^8$. Using the binomial expansion,$(x+y)^8 + (x-y)^8 = 2 \sum_{k=0, 2, 4, 6, 8} \binom{8}{k} x^{8-k} y^k$. Substituting $y^2 = x^2-1$,the terms are: $2 [ \binom{8}{0} x^8 + \binom{8}{2} x^6(x^2-1) + \binom{8}{4} x^4(x^2-1)^2 + \binom{8}{6} x^2(x^2-1)^3 + \binom{8}{8} (x^2-1)^4 ]$. The highest power of $x$ is $x^8$. The coefficient of $x^8$ is $2 [ \binom{8}{0} + \binom{8}{2} + \binom{8}{4} + \binom{8}{6} + \binom{8}{8} ]$. We know that $\sum_{k 	ext{ even}} \binom{n}{k} = 2^{n-1}$. Thus,the sum is $2 	imes 2^{8-1} = 2 	imes 2^7 = 2^8 = 256$.

The coefficient of the highest power of $x$ in the expansion of $(x+\sqrt{x^2-1})^8+(x-\sqrt{x^2-1})^8$ is

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