If the ratio of the terms equidistant from th | vedclass

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(B) The middle term is the $(\frac{12}{2}+1)^{	ext{th}}$ term,i.e.,the $7^{	ext{th}}$ term. The terms equidistant from the middle term are $T_{7-k}$

Vedclass · Accepted Answer

$3^{12}$ or $5^{12}$

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(B) The middle term is the $(\frac{12}{2}+1)^{	ext{th}}$ term,i.e.,the $7^{	ext{th}}$ term. The terms equidistant from the middle term are $T_{7-k}$ and $T_{7+k}$. For $k=2$,the ratio is $\frac{T_5}{T_9} = \frac{{}^{12}C_4 x^4}{{}^{12}C_8 x^8} = \frac{1}{256}$. Since ${}^{12}C_4 = {}^{12}C_8$,we have $\frac{1}{x^4} = \frac{1}{256}$ $\Rightarrow x^4 = 256$ $\Rightarrow x = 4$. For $k=4$,the ratio is $\frac{T_3}{T_{11}} = \frac{{}^{12}C_2 x^2}{{}^{12}C_{10} x^{10}} = \frac{1}{256}$. Since ${}^{12}C_2 = {}^{12}C_{10}$,we have $\frac{1}{x^8} = \frac{1}{256}$ $\Rightarrow x^8 = 256$ $\Rightarrow x = \sqrt{2}$. However,given $x \in N$,we check $k=1$: $\frac{T_6}{T_8} = \frac{{}^{12}C_5 x^5}{{}^{12}C_7 x^7} = \frac{1}{x^2} = \frac{1}{256} \Rightarrow x = 16$. The sum of all terms in the expansion $(1+x)^{12}$ is $(1+x)^{12}$. For $x=4$,the sum is $(1+4)^{12} = 5^{12}$. For $x=2$,the sum is $(1+2)^{12} = 3^{12}$.

If the ratio of the terms equidistant from the middle term in the expansion of $(1+x)^{12}$ is $\frac{1}{256}$ $(x \in N)$,then the sum of all the terms of the expansion $(1+x)^{12}$ is:

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