Question

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

Answer 1

To prove that the numbers 49, 4489, 444889, and so on, obtained by inserting 48 into the middle of the preceding number, are squares of integers, we can use mathematical induction. By assuming the inductive hypothesis is true for some k and showing that it is also true for (k + 1), we can conclude that all the numbers in the sequence are squares of integers.

Step-by-step explanation:

To prove that the numbers 49, 4489, 444889, and so on, obtained by inserting 48 into the middle of the preceding number, are squares of integers, we can use mathematical induction.

1. Base Case:

   n = 1: The first number is 49, which is equal to 7². Hence, the base case is true.

2. Inductive Hypothesis:

   Assume that for some positive integer k, the number 44...48...88 is equal to m², where m is an integer.

3. Inductive Step:

   We need to prove that the number 44...48...88, formed by inserting 48 into the middle of the preceding number (k + 1) times, is equal to a perfect square.

   Let's begin by expressing the number in terms of the inductive hypothesis:

   44...48...88 = 44...44 + 4(48) + 44...44

   Since the inductive hypothesis assumes that the previous number (k times) is a perfect square, we can write it as (m²).

   Now, let's write the number as a square:

   44...48...88 = (m²)(10^(k+1)) + 4(48) + (m²)

   Simplifying further, we get:

   44...48...88 = (10^(k+1) + 1)(m²) + 4(48)

   Now, let's express the number as a perfect square:

   44...48...88 = (10^(k+1) + 1)(m²) + 2²(2² x 24)

   44...48...88 = (10^(k+1) + 1)(m²) + 2²(4² x 6)

   44...48...88 = (10^(k+1) + 1)(m²) + (4²)(2² x 6)

   44...48...88 = (10^(k+1) + 1)(m² + (4²)(2² x 6))

   44...48...88 = (10^(k+1) + 1)(m² + 384)

   Thus, the number formed by inserting 48 into the middle of the preceding number (k + 1) times is equal to the perfect square (10^(k+1) + 1)(m² + 384).

   Since the base case is true, and assuming the inductive hypothesis is true for some k, we have shown that it is also true for (k + 1).

   Therefore, by mathematical induction, we can conclude that the numbers 49, 4489, 444889, and so on, obtained by inserting 48 into the middle of the preceding number, are all squares of integers.