If the ratio of the terms equidistant from th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $5^{12}$

Vedclass · Answer

(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:

Similar Questions

Let $m$ be the smallest positive integer such that the coefficient of $x^2$ in the expansion of $(1+x)^2+(1+x)^3+\cdots+(1+x)^{49}+(1+mx)^{50}$ is $(3n+1)^{51}C_3$ for some positive integer $n$. Then the value of $n$ is

Write the general term in the expansion of $(x^{2}-yx)^{12}, x \neq 0$.

If $A$ and $B$ are coefficients of $x^{n}$ in the expansions of $(1+x)^{2n}$ and $(1+x)^{2n-1}$ respectively, then $A / B$ is equal to

The value of $x$,for which the $6^{th}$ term in the expansion of ${\left\{ {{2^{{{\log }_2}\sqrt {({9^{x - 1}} + 7)} }} + \frac{1}{{{2^{(1/5){{\log }_2}({3^{x - 1}} + 1)}}}}} \right\}^7}$ is $84$,is equal to

If the middle term in the expansion of ${\left( {x^2 + \frac{1}{x}} \right)^n}$ is $924x^6$,then $n = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module