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Question: The sum of \[12\] term of the series: \({}^{12}{C_1} \cdot \dfrac{1}{3} + {}^{12}{C_2} \cdot \dfrac{...

The sum of 1212 term of the series: 12C1⋅13+12C2⋅19+12C3⋅127+........{}^{12}{C_1} \cdot \dfrac{1}{3} + {}^{12}{C_2} \cdot \dfrac{1}{9} + {}^{12}{C_3} \cdot \dfrac{1}{{27}} + ........ is-
(A) (43)12−1{\left( {\dfrac{4}{3}} \right)^{12}} - 1
(B) (34)12−1{\left( {\dfrac{3}{4}} \right)^{12}} - 1
(C) (34)12+1{\left( {\dfrac{3}{4}} \right)^{12}} + 1
(D)none of these

Explanation

Solution

Use the formula (1+x)n=nC0(x)0+nC1(x)1+nC2(x)2+...........+nCn(x)n{\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 1212 term of the series: 12C1⋅13+12C2⋅19+12C3⋅127+........{}^{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 1212 term of the series: 12C1⋅13+12C2⋅19+12C3⋅127+........{}^{12}{C_1} \cdot \dfrac{1}{3} + {}^{12}{C_2} \cdot \dfrac{1}{9} + {}^{12}{C_3} \cdot \dfrac{1}{{27}} + .........
We know that, (1+x)n=nC0(x)0+nC1(x)1+nC2(x)2+...........+nCn(x)n{\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=12n = 12 and x=13x = \dfrac{1}{3}, we get-
(1+13)12=12C0(13)0+12C1(13)1+12C2(13)2+...........+12C12(13)12{\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 (3+13)12=12C0+12C1(13)+12C2(19)+...........+12C12(13)12{\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 nC0=1{}^n{C_0} = 1, therefore, 12C0=1{}^{12}{C_0} = 1
⇒\Rightarrow (43)12=1+12C1(13)+12C2(19)+...........+12C12(13)12{\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 12C1(13)+12C2(19)+...........+12C12(13)12=(43)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}} = {\left( {\dfrac{4}{3}} \right)^{12}} - 1

Hence, option (A) is the correct answer.

Note: The value of 12C0{}^{12}{C_0} can be calculate as follows:
We know that nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
Put n=12,r=0n = 12,r = 0
12C0=12!0!(12−0)!{}^{12}{C_0} = \dfrac{{12!}}{{0!\left( {12 - 0} \right)!}}
12C0=12!12!{}^{12}{C_0} = \dfrac{{12!}}{{12!}}
12C0=1{}^{12}{C_0} = 1