If T_4 represents the 4^{th} term in the expa | vedclass

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(C) We know the expansion of $(1+y)^n = 1 + ny + \frac{n(n-1)y^2}{2!} + \frac{n(n-1)(n-2)y^3}{3!} + \dots$ Given expression: $\left(5x + \frac{7}{x}\r

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$-\frac{7^4}{2^4 5^3}$

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(C) We know the expansion of $(1+y)^n = 1 + ny + \frac{n(n-1)y^2}{2!} + \frac{n(n-1)(n-2)y^3}{3!} + \dots$ Given expression: $\left(5x + \frac{7}{x}ight)^{-3/2} = (5x)^{-3/2} \left(1 + \frac{7}{5x^2}ight)^{-3/2}$. Let $y = \frac{7}{5x^2}$ and $n = -3/2$. The $4^{th}$ term $T_4$ is given by the term containing $y^3$: $T_4 = (5x)^{-3/2} 	imes \frac{n(n-1)(n-2)}{3!} y^3$. Substituting the values: $T_4 = (5x)^{-3/2} 	imes \frac{(-3/2)(-5/2)(-7/2)}{6} \left(\frac{7}{5x^2}ight)^3$. $T_4 = 5^{-3/2} x^{-3/2} 	imes \left(-\frac{105}{48}ight) 	imes \frac{7^3}{5^3 x^6} = 5^{-3/2} x^{-3/2} 	imes \left(-\frac{35}{16}ight) 	imes \frac{7^3}{5^3 x^6}$. $T_4 = -\frac{35 	imes 7^3}{16 	imes 5^{3/2} 	imes 5^3 	imes x^{15/2}} = -\frac{5 	imes 7^4}{2^4 	imes 5^{9/2} x^{15/2}} = -\frac{7^4}{2^4 	imes 5^{7/2} x^{15/2}}$. Now,calculate $\left(x^7 \sqrt{5x}ight) T_4 = x^7 \cdot 5^{1/2} x^{1/2} \cdot \left(-\frac{7^4}{2^4 \cdot 5^{7/2} x^{15/2}}ight)$. $= -\frac{7^4}{2^4 \cdot 5^{7/2 - 1/2} \cdot x^{15/2 - 15/2}} = -\frac{7^4}{2^4 \cdot 5^3}$.

If $T_4$ represents the $4^{th}$ term in the expansion of $\left(5x + \frac{7}{x}\right)^{-3/2}$ and $x \notin \left[-\sqrt{\frac{7}{5}}, \sqrt{\frac{7}{5}}\right]$,then $\left(x^7 \sqrt{5x}\right) T_4 =$

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