Question

The sum of \[12\] term of the series: ${}^{12}{C_1} \cdot \dfrac{1}{3} + {}^{12}{C_2} \cdot \dfrac{1}{9} + {}^{12}{C_3} \cdot \dfrac{1}{{27}} + ........$ is-
(A) ${\left( {\dfrac{4}{3}} \right)^{12}} - 1$
(B) ${\left( {\dfrac{3}{4}} \right)^{12}} - 1$
(C) ${\left( {\dfrac{3}{4}} \right)^{12}} + 1$
(D)none of these

Answer 1

Hint: Use the formula ${\left( {1 + x} \right)^n} = {}^n{C_0}{\left( x \right)^0} + {}^n{C_1}{\left( x \right)^1} + {}^n{C_2}{\left( x \right)^2} + ........... + {}^n{C_n}{\left( x \right)^n}$ to find the sum of \[12\] term of the series: ${}^{12}{C_1} \cdot \dfrac{1}{3} + {}^{12}{C_2} \cdot \dfrac{1}{9} + {}^{12}{C_3} \cdot \dfrac{1}{{27}} + ........$

Complete step-by-step answer:
We have to find the sum of \[12\] term of the series: ${}^{12}{C_1} \cdot \dfrac{1}{3} + {}^{12}{C_2} \cdot \dfrac{1}{9} + {}^{12}{C_3} \cdot \dfrac{1}{{27}} + ........$.
 We know that, ${\left( {1 + x} \right)^n} = {}^n{C_0}{\left( x \right)^0} + {}^n{C_1}{\left( x \right)^1} + {}^n{C_2}{\left( x \right)^2} + ........... + {}^n{C_n}{\left( x \right)^n}$
Put $n = 12$ and $x = \dfrac{1}{3}$, we get-
${\left( {1 + \dfrac{1}{3}} \right)^{12}} = {}^{12}{C_0}{\left( {\dfrac{1}{3}} \right)^0} + {}^{12}{C_1}{\left( {\dfrac{1}{3}} \right)^1} + {}^{12}{C_2}{\left( {\dfrac{1}{3}} \right)^2} + ........... + {}^{12}{C_{12}}{\left( {\dfrac{1}{3}} \right)^{12}}$
$ \Rightarrow $${\left( {\dfrac{{3 + 1}}{3}} \right)^{12}} = {}^{12}{C_0} + {}^{12}{C_1}\left( {\dfrac{1}{3}} \right) + {}^{12}{C_2}\left( {\dfrac{1}{9}} \right) + ........... + {}^{12}{C_{12}}{\left( {\dfrac{1}{3}} \right)^{12}}$
We know that ${}^n{C_0} = 1$, therefore, ${}^{12}{C_0} = 1$
$ \Rightarrow $${\left( {\dfrac{4}{3}} \right)^{12}} = 1 + {}^{12}{C_1}\left( {\dfrac{1}{3}} \right) + {}^{12}{C_2}\left( {\dfrac{1}{9}} \right) + ........... + {}^{12}{C_{12}}{\left( {\dfrac{1}{3}} \right)^{12}}$
$ \Rightarrow $\[{}^{12}{C_1}\left( {\dfrac{1}{3}} \right) + {}^{12}{C_2}\left( {\dfrac{1}{9}} \right) + ........... + {}^{12}{C_{12}}{\left( {\dfrac{1}{3}} \right)^{12}} = {\left( {\dfrac{4}{3}} \right)^{12}} - 1\]

Hence, option (A) is the correct answer.

Note: The value of ${}^{12}{C_0}$ can be calculate as follows:
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Put $n = 12,r = 0$
${}^{12}{C_0} = \dfrac{{12!}}{{0!\left( {12 - 0} \right)!}}$
${}^{12}{C_0} = \dfrac{{12!}}{{12!}}$
${}^{12}{C_0} = 1$