The expression frac{1}{{sqrt {4x + 1} }}left[ | vedclass

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(D) Let $y = \sqrt{4x + 1}$. The expression becomes $\frac{1}{y} \left[ \left( \frac{1+y}{2} \right)^7 - \left( \frac{1-y}{2} \right)^7 \right]$.Using

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$3$

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(D) Let $y = \sqrt{4x + 1}$. The expression becomes $\frac{1}{y} \left[ \left( \frac{1+y}{2} ight)^7 - \left( \frac{1-y}{2} ight)^7 ight]$.Using the binomial expansion,$\left( \frac{1+y}{2} ight)^7 = \frac{1}{2^7} \sum_{k=0}^7 \binom{7}{k} y^k$ and $\left( \frac{1-y}{2} ight)^7 = \frac{1}{2^7} \sum_{k=0}^7 \binom{7}{k} (-y)^k$.The difference is $\frac{1}{2^7} \sum_{k=0}^7 \binom{7}{k} (y^k - (-y)^k)$.Terms with even $k$ cancel out,leaving only odd $k$ terms: $2 	imes \frac{1}{2^7} \left( \binom{7}{1} y + \binom{7}{3} y^3 + \binom{7}{5} y^5 + \binom{7}{7} y^7 ight)$.Dividing by $y$,we get $\frac{2}{2^7} \left( \binom{7}{1} + \binom{7}{3} y^2 + \binom{7}{5} y^4 + \binom{7}{7} y^6 ight)$.Since $y^2 = 4x + 1$,the highest power of $x$ is $(y^2)^3 = (4x+1)^3$,which is $x^3$.Thus,the degree of the polynomial is $3$.

The expression $\frac{1}{{\sqrt {4x + 1} }}\left[ {{{\left[ {\frac{{1 + \sqrt {4x + 1} }}{2}} \right]}^7} - {{\left[ {\frac{{1 - \sqrt {4x + 1} }}{2}} \right]}^7}} \right]$ is a polynomial in $x$ of degree

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