If the ratio of the terms equidistant from the middle term in the expansion of $(1+x)^{12}$ is…

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 sum of all the terms of the expansion $(1+x)^{12}$ is
  1. $4^{12}$ or $6^{12}$
  2. $3^{12}$ or $5^{12}$
  3. $6^{12}$ or $7^{12}$
  4. $12^{12}$

Solution

Middle term $\left(\frac{n}{2}+1\right)^{\text {th }}$ term i.e., $7^{\text {th }}$ term $T_7={ }^{12} C_6 x^6$
Now, $\frac{{ }^{12} C_8 x^4}{{ }^{12} C_4 x^8}=\frac{1}{256} \Rightarrow x=4$
and $\frac{{ }^{12} C_{10} x^2}{{ }^{12} C_2 x^{10}}=\frac{1}{256} \Rightarrow x=2$ $\therefore$ Sum of all term $=(1+4)^{12}$ or $(1+2)^{12}=5^{12}$ or $3^{12}$.

Asked in: AP EAMCET 2024 (21 May Shift 2)

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