Question

The numbers 49, 4489, 444889 ... obtained by inserting 48 into the middle of the preceding square numbers are square of integers.

Answer 1

Hint: Every n-digit number in decimal (base 10) is of the form $ {{a}_{n-1}}\times {{10}^{n-1}}+...+{{a}_{1}}\times {{10}^{1}}+{{a}_{0}}\times {{10}^{0}} $ , where $ {{a}_{n-1}},...,{{a}_{1}},{{a}_{0}} $ are some digits from 0 to 9.
Express the general n-digit number of the given type in terms of number of 4s, 8s, 9s and n.
Use the formula for the sum of a geometric series:
 $ {{S}_{n}}=a\left( \dfrac{{{r}^{n}}-1}{r-1} \right) $ , where a is the first term, r is the common ratio and n is the number of terms.

Complete step-by-step answer:
The number formed using the given conditions will have an even number of digits.
Let us consider the $ 2n+2 $ digit number 444...8889.
There are half as many 4s and half as many 8s but one 9.
The number of 4s will be $ n+1 $ , 8s will be $ n $ and 9s will be just 1.
 $ \underbrace{4444...44}_{n+1}\underbrace{888...88}_{n}\underbrace{9}_{1} $
The number can be written in expanded form as:
$ 4\times \left( {{10}^{2n+1}}+{{10}^{2n}}+...+{{10}^{n+1}} \right)+8\times \left( {{10}^{n}}+{{10}^{n-1}}+...+{{10}^{1}} \right)+9\times {{10}^{0}} $
Since the sums of powers of 10 form a geometric series, we can write it as:
= $ 4\times \left[ {{10}^{n+1}}\left( \dfrac{{{10}^{n+1}}-1}{10-1} \right) \right]+8\times \left[ 10\left( \dfrac{{{10}^{n}}-1}{10-1} \right) \right]+9 $
Separating $ \dfrac{1}{9} $ as a common factor from all the terms, we get:
= $ \dfrac{1}{9}\left[ 4\times {{10}^{n+1}}\left( {{10}^{n+1}}-1 \right)+8\times 10\left( {{10}^{n}}-1 \right)+81 \right] $
On multiplying the terms inside the bracket, we get:
= $ \dfrac{1}{9}\left( 4\times {{10}^{2n+2}}-4\times {{10}^{n+1}}+8\times {{10}^{n+1}}-80+81 \right) $
Let us put $ {{10}^{n+1}}=x $ for easier writing, observation and manipulation.
= $ \dfrac{1}{9}\left( 4{{x}^{2}}+4x+1 \right) $
= $ \dfrac{1}{9}{{\left( 2x+1 \right)}^{2}} $
= $ {{\left( \dfrac{2x+1}{3} \right)}^{2}} $
= $ {{\left( \dfrac{2\times {{10}^{n+1}}+1}{3} \right)}^{2}} $
Since the expression $ 2\times {{10}^{n+1}}+1 $ is always divisible by 3 for any value of n (the sum of the digits is always 3), we can say that the given number is the square of an integer, for any positive integer value of n.
Hence proved.

Note: The expanded form of a number is more useful in solving equations as it clearly shows the mathematical relation between the individual numbers (digits).
e.g. A two-digit number can be written as $ 10x+y $ , which is a more usable form in equations.
The number formed by recurring decimals is the sum of an infinite series whose common ratio is of the form $ \dfrac{1}{{{10}^{n}}} $ .
There are many number bases (like binary, hexa-decimal, etc.) used in various applications, but base 10 is the most widely used form since ancient times, chiefly because we have 10 fingers to count.