If t_r is the r^{text{th}} term in the expans | vedclass

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(D) The general term in the expansion of $(a + b)^n$ is $t_{r+1} = \binom{n}{r} a^{n-r} b^r$.For the expansion $(2^x + 4^{-x})^8$,we have $n=8$,$a=2^x

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$-1/3$

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(D) The general term in the expansion of $(a + b)^n$ is $t_{r+1} = \binom{n}{r} a^{n-r} b^r$.For the expansion $(2^x + 4^{-x})^8$,we have $n=8$,$a=2^x$,and $b=4^{-x} = (2^2)^{-x} = 2^{-2x}$.$t_2 = t_{1+1} = \binom{8}{1} (2^x)^7 (2^{-2x})^1 = 8 \cdot 2^{7x} \cdot 2^{-2x} = 8 \cdot 2^{5x}$.$t_3 = t_{2+1} = \binom{8}{2} (2^x)^6 (2^{-2x})^2 = 28 \cdot 2^{6x} \cdot 2^{-4x} = 28 \cdot 2^{2x}$.Given $t_3 = 7t_2$,we substitute the expressions:$28 \cdot 2^{2x} = 7 \cdot (8 \cdot 2^{5x})$.$28 \cdot 2^{2x} = 56 \cdot 2^{5x}$.Divide both sides by $28 \cdot 2^{2x}$:$1 = 2 \cdot 2^{3x} = 2^{3x+1}$.Since $1 = 2^0$,we have $3x + 1 = 0$,which gives $x = -1/3$.

If $t_r$ is the $r^{\text{th}}$ term in the expansion of $(2^x + 4^{-x})^8$ and if $t_3 = 7t_2$,then $x =$

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