Question

If \[x \in \left\{ {1,2,3,.....9} \right\}\] and \[{f_n}(x) = xxx.....x\](n digits), then \[{f_n}^2(3) + {f_n}(2)\] is equal to
A. \[2{f_{2n}}(1)\]
B. \[{f_n}^2(1)\]
C. \[{f_{2n}}(1)\]
D. \[ - {f_{2n}}(4)\]

Answer 1

Hint: Here we form the first few terms using substitution of values of n in the function. Write the terms after breaking them using place values. Calculate the sum of the geometric progression formed and write the value of the terms required in the equation. Calculate the value of the equation and check which of the given options match the answer.
* A geometric progression is a sequence of terms where each next term can be calculated by multiplying the term with the common ratio. If a GP has terms \[a,{a_2},{a_3}......{a_n}\] having common ratio ‘r’ then sum of ‘n’ terms of GP is given by \[S = \dfrac{{a({r^n} - 1)}}{{(r - 1)}}\]

Complete step-by-step solution:
Since we have \[{f_n}(x) = xxx.....x\](n digits)
If we put the value of \[n = 1 \Rightarrow {f_1}(x) = x\]........................… (1)
If we put the value of \[n = 2 \Rightarrow {f_2}(x) = xx\]
We can break the value\[{f_2}(x)\]using ones and tens place.
\[\because xx = x + 10x\]
\[ \Rightarrow {f_2}(x) = x + 10x\].......................… (2)
If we put the value of \[n = 3 \Rightarrow {f_3}(x) = xxx\]
We can break the value\[{f_3}(x)\]using ones, tens and hundreds place.
\[\because xxx = x + 10x + 100x\]
i.e. \[xxx = x + 10x + {10^2}x\]
\[ \Rightarrow {f_3}(x) = x + 10x + {10^2}x\]........................… (3)
So we can write general equation for ‘n’
\[ \Rightarrow {f_n}(x) = x + 10x + {10^2}x + {......10^{n - 1}}x\]..................… (4)
Now we take ‘x’ common from all terms in RHS of the equation (4)
\[ \Rightarrow {f_n}(x) = x\left( {1 + 10 + {{10}^2} + {{......10}^{n - 1}}} \right)\].............… (5)
The terms in the bracket form a GP as common ratio between each term is 10 i.e. \[\dfrac{{10}}{1} = 10;\dfrac{{100}}{{10}} = 10;......\dfrac{{{{10}^{n - 1}}}}{{{{10}^{n - 2}}}} = 10\]
So, the GP has a first term as 1, common ratio as 10. Use the formula of sum of n terms of a GP
\[ \Rightarrow {S_n} = \dfrac{{1({{10}^n} - 1)}}{{(10 - 1)}}\]
\[ \Rightarrow {S_n} = \dfrac{{{{10}^n} - 1}}{9}\]
So the value of \[\left( {1 + 10 + {{10}^2} + {{......10}^{n - 1}}} \right)\] is \[\dfrac{{{{10}^n} - 1}}{9}\].
Substitute the value of \[\left( {1 + 10 + {{10}^2} + {{......10}^{n - 1}}} \right)\] in equation (5)
\[ \Rightarrow {f_n}(x) = \dfrac{{x\left( {{{10}^n} - 1} \right)}}{9}\]..................… (6)
Now we have to find the value of the equation\[{f_n}^2(3) + {f_n}(2)\].
Calculate \[{f_n}(3)\] and \[{f_n}(2)\] separately.
Substitute value of \[x = 3\]in equation (6)
\[ \Rightarrow {f_n}(3) = \dfrac{{3\left( {{{10}^n} - 1} \right)}}{9}\]
Cancel same terms from numerator and denominator
\[ \Rightarrow {f_n}(3) = \dfrac{{\left( {{{10}^n} - 1} \right)}}{3}\]
Square both sides of the equation
\[ \Rightarrow {f_n}^2(3) = \dfrac{{{{\left( {{{10}^n} - 1} \right)}^2}}}{{{3^2}}}\]
\[ \Rightarrow {f_n}^2(3) = \dfrac{{{{\left( {{{10}^n} - 1} \right)}^2}}}{9}\].................… (7)
Substitute value of \[x = 2\]in equation (6)
\[ \Rightarrow {f_n}(2) = \dfrac{{2\left( {{{10}^n} - 1} \right)}}{9}\]...................… (8)
Substitute values from equations (7) and (8) in the equation\[{f_n}^2(3) + {f_n}(2)\]
\[ \Rightarrow {f_n}^2(3) + {f_n}(2) = \dfrac{{{{\left( {{{10}^n} - 1} \right)}^2}}}{9} + \dfrac{{2\left( {{{10}^n} - 1} \right)}}{9}\]
Take \[\dfrac{{\left( {{{10}^n} - 1} \right)}}{9}\]common in RHS
\[ \Rightarrow {f_n}^2(3) + {f_n}(2) = \dfrac{{\left( {{{10}^n} - 1} \right)}}{9}\left( {\left( {{{10}^n} - 1} \right) + 2} \right)\]
Calculate the value in the bracket
\[ \Rightarrow {f_n}^2(3) + {f_n}(2) = \dfrac{{\left( {{{10}^n} - 1} \right)}}{9}\left( {{{10}^n} - 1 + 2} \right)\]
\[ \Rightarrow {f_n}^2(3) + {f_n}(2) = \dfrac{{\left( {{{10}^n} - 1} \right)}}{9}\left( {{{10}^n} + 1} \right)\]
Use the identity \[(a - b)(a + b) = {a^2} - {b^2}\]
\[ \Rightarrow {f_n}^2(3) + {f_n}(2) = \dfrac{{\left( {{{({{10}^n})}^2} - {{(1)}^2}} \right)}}{9}\]
\[ \Rightarrow {f_n}^2(3) + {f_n}(2) = \dfrac{{\left( {{{10}^{2n}} - 1} \right)}}{9}\] ……………..… (9)
Now we check for the options
Since the power of 10 in the answer is 2n, we will consider the terms having subscript 2n first
Calculate \[{f_{2n}}(1)\]
Since \[{f_n}(x) = \dfrac{{x\left( {{{10}^n} - 1} \right)}}{9}\]
\[ \Rightarrow {f_{2n}}(x) = \dfrac{{x\left( {{{10}^{2n}} - 1} \right)}}{9}\]
Put \[x = 1\]in the above equation
\[ \Rightarrow {f_{2n}}(1) = \dfrac{{1\left( {{{10}^{2n}} - 1} \right)}}{9}\]
\[ \Rightarrow {f_{2n}}(1) = \dfrac{{\left( {{{10}^{2n}} - 1} \right)}}{9}\].......................… (10)
Since RHS of equations (9) and (10) are equal then LHS of equations are also equal.
\[ \Rightarrow {f_n}^2(3) + {f_n}(2) = {f_{2n}}(1)\]
\[\therefore \]The value of \[{f_n}^2(3) + {f_n}(2)\] is equal to\[{f_{2n}}(1)\]

\[\therefore \]Correct option is C.

Note: Many students get confused while writing the initial values of the functions as they write \[xx = {x^2};xxx = {x^3}...\]. This is the wrong approach as we are given that the function gives us n digits of the same type, so numbers formed can be like 999, 44, 333333 etc.