If the fourth term in the expansion of (x+x^{ | vedclass

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(A) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.For the fourth term $(T_4)$,we set $r=3

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$2$

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(A) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.For the fourth term $(T_4)$,we set $r=3$:$T_4 = {}^{7}C_{3} (x)^{7-3} (x^{\log_{2} x})^{3} = 4480$.Since ${}^{7}C_{3} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$,we have:$35 \cdot x^{4} \cdot x^{3 \log_{2} x} = 4480$.Dividing by $35$:$x^{4 + 3 \log_{2} x} = \frac{4480}{35} = 128 = 2^{7}$.Let $t = \log_{2} x$,then $x = 2^t$. Substituting this into the equation:$(2^t)^{4 + 3t} = 2^{7} \Rightarrow t(4 + 3t) = 7$.$3t^{2} + 4t - 7 = 0$.$(3t + 7)(t - 1) = 0$.Thus,$t = 1$ or $t = -7/3$.If $t = 1$,then $\log_{2} x = 1 \Rightarrow x = 2^1 = 2$.Since $x \in N$,the value is $2$.

If the fourth term in the expansion of $(x+x^{\log _{2} x})^{7}$ is $4480$,then the value of $x$ where $x \in N$ is equal to

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