Question

What is the value of \[\left( {\dfrac{1}{{8{1^n}}}} \right) - \left( {\dfrac{{10}}{{81^{n}}}} \right){}^{2n}{C_1} + \left( {\dfrac{{10^{2}}}{{81^{n}}}} \right){}^{2n}{C_2} - \left( {\dfrac{{10^{3}}}{{81^{n}}}} \right){}^{2n}{C_3} + .... + \left( {\dfrac{{10^{2n}}}{{81^{n}}}} \right)\]?
A. 2
B. 0
C. \[\dfrac{1}{2}\]
D. 1

Answer 1

Hint: First we will take common \[\left( {\dfrac{1}{{8{1^n}}}} \right)\] from the given expression \[\left( {\dfrac{1}{{8{1^n}}}} \right) - \left( {\dfrac{{10}}{{8{1^n}}}} \right){}^{2n}{C_1} + \left( {\dfrac{{1{0^2}}}{{8{1^n}}}} \right){}^{2n}{C_2} - \left( {\dfrac{{1{0^3}}}{{8{1^n}}}} \right){}^{2n}{C_3} + .... + \left( {\dfrac{{1{0^{2n}}}}{{8{1^n}}}} \right)\]. Then we apply reverse binomial theorem that is \[1 - {}^n{C_1}x + {}^n{C_2}{x^2} + \cdots \cdots + {}^n{C_n}{x^n} = {\left( {1 - x} \right)^n}\]. Then simplify the expression to get the value of the given expression.

Formula used :
Binomial expansion of \[{\left( {1 - x} \right)^n}\]:
\[{\left( {1 - x} \right)^n} = 1 - nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} - \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + .....\]
\[{a^{mn}} = {\left( {{a^m}} \right)^n}\]

Complete step by step solution:
The given expression is \[\left( {\dfrac{1}{{81^{n}}}} \right) - \left( {\dfrac{{10}}{{81^{n}}}} \right){}^{2n}{C_1} + \left( {\dfrac{{10^{2}}}{{81^{n}}}} \right){}^{2n}{C_2} - \left( {\dfrac{{10^{3}}}{{81^{n}}}} \right){}^{2n}{C_3} + .... + \left( {\dfrac{{10^{2n}}}{{81^{n}}}} \right)\].
Let’s simplify the given expression.
Let \[V\] be the value of the above expression.
\[V = \left( {\dfrac{1}{{81^{n}}}} \right) - \left( {\dfrac{{10}}{{81^{n}}}} \right){}^{2n}{C_1} + \left( {\dfrac{{10^{2}}}{{81^{n}}}} \right){}^{2n}{C_2} - \left( {\dfrac{{10^{3}}}{{81^{n}}}} \right){}^{2n}{C_3} + .... + \left( {\dfrac{{10^{2n}}}{{81^{n}}}} \right)\]
Factor out the common factor from each term.
\[V = \dfrac{1}{{81^{n}}}\left[ {1 - 10 \cdot {}^{2n}{C_1} + 10^{2} \cdot {}^{2n}{C_2} - 10^{3} \cdot {}^{2n}{C_3} + .... + 10^{2n}} \right]\]
Now apply the reverse formula of binomial expansion \[{\left( {1 - x} \right)^n}\].
The terms present in the square bracket are the binomial expansion of \[{\left( {1 - 10} \right)^{2n}}\].
Then,
\[V = \dfrac{1}{{81^{n}}}{\left[ {1 - 10} \right]^{2n}}\]
\[V = \dfrac{1}{{81^{n}}}{\left[ { - 9} \right]^{2n}}\]
Since \[2n\] is an even number and the value of even power of any negative number is positive.
So, \[{\left[ { - 9} \right]^{2n}} = {\left[ 9 \right]^{2n}}\]
\[V = \dfrac{1}{{81^{n}}}{\left[ 9 \right]^{2n}}\]
Now apply the property of powers \[{a^{mn}} = {\left( {{a^m}} \right)^n}\].
\[V = \dfrac{{{{\left[ {{9^2}} \right]}^n}}}{{81^{n}}}\]
\[ \Rightarrow \]\[V = \dfrac{{81^{n}}}{{81^{n}}}\] [ Since \[{9^2} = 81\]]
\[ \Rightarrow \]\[V = 1\]
Hence the correct option is D

Note: Students often do a common mistake that is \[{\left( { - 9} \right)^{2n}} = - {81^n}\] which is wrong. Since \[2n\] is an even number. So the value of \[{\left( { - 9} \right)^{2n}}\] must be a positive number.