Question

If \[{\left( {666... n times} \right)^2} + \left( {8888... n times} \right) = \left( {4444... K times} \right)\], then find the value of \[K\].
A. \[n + 1\]
B. \[n\]
C. \[2n\]
D. \[{n^2}\]

Answer 1

Hint In the given question, an equation is given. First, simplify the left-hand side of the given equation. The terms of the equation are in geometric progression. So, by using the formula of the sum of \[n\] terms in geometric progression, simplify the left-hand side of the given equation. Then compare the left-hand side with the right-hand side of the equation.

Formula used
The sum of \[n\] terms in geometric progression: \[{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}\] , if \[r \ne 1\] and \[r > 1\].
    where \[a\] is the first term, \[r\] is the common ratio and \[n\] is the number of terms
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]

Complete step by step solution:
The given equation is \[{\left( {666... n times} \right)^2} + \left( {8888... n times} \right) = \left( {4444 K times} \right)\]
Simplify the left-hand side of the equation.
\[{\left( {666... n times} \right)^2} + \left( {8888... n times} \right)\]
\[= {\left( {6 + 60 + 600 + ... n times} \right)^2} + \left( {8 + 80 + 800 + ... n times} \right)\]
\[= 36{\left( {1 + 10 + 100 + ... n times} \right)^2} + 8\left( {1 + 10 + 100 + ... n times} \right)\]
The terms present in both brackets are in geometric progression where the first term is \[a = 1\] and the common ratio is \[r = 10\].
Apply the formula of the sum of \[n\] terms in geometric progression in the brackets.
\[ = 36{\left( {\dfrac{{1\left( {1{0^n} - 1} \right)}}{{10 - 1}}} \right)^2} + 8\left( {\dfrac{{1\left( {1{0^n} - 1} \right)}}{{10 - 1}}} \right)\]
Simplify the above expression.
\[= 36{\left( {\dfrac{{1{0^n} - 1}}{9}} \right)^2} + 8\left( {\dfrac{{1{0^n} - 1}}{9}} \right)\]
Apply the formula \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\].
\[ = 36\left( {\dfrac{{1{0^{2n}} + 1 - 2 \cdot 1{0^n}}}{{81}}} \right) + 8\left( {\dfrac{{1{0^n} - 1}}{9}} \right)\]
\[ = \dfrac{4}{9}\left( {1{0^{2n}} + 1 - 2 \times 1{0^n} + 2 \times 1{0^n} - 2} \right)\]
\[ = \dfrac{4}{9}\left( {1{0^{2n}} - 1} \right)\]
\[ = 4\left( {\dfrac{{1{0^{2n}} - 1}}{{10 - 1}}} \right)\]
Using the reverse formula of the sum of the geometric series.
\[ = 4\left( {1 + 10 + 100 + .... 2n times} \right)\]
\[ = 4 + 40 + 400 + .... 2n times\]
\[ = 4444.... 2n times\]
Thus,
\[{\left( {666... n times} \right)^2} + \left( {8888... n times} \right) = \left( {4444... 2n times} \right)\] \[.....equation\left( 1 \right)\]
The given equation is,
 \[{\left( {666... n times} \right)^2} + \left( {8888... n times} \right) = \left( {4444... K times} \right)\] \[.....equation\left( 2 \right)\]
Equate the right-hand side of the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
Then,
\[K = 2n\]
Hence the correct option is option C.

Note: Students are often confused with the formula of the sum of the \[n\] terms in geometric progression.
If the common ratio \[r \ne 1\] and \[r > 1\], then the sum of the \[n\] terms in GP is: \[{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}\]
If the common ratio \[r \ne 1\] and \[r < 1\], then the sum of the \[n\] terms in GP is: \[{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}\]
where \[a\] is the first term, \[r\] is the common ratio and \[n\] is the number of terms.