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

1
by karthik4297

2014-10-23T05:58:08+05:30

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Let the number N consist of 2m digits.
m = 1,  N = 49       m =2    N = 4489       m = 3    N = 444889
It can be written as
m = 1,  N-1 = 48       m =2    N-1 = 4488       m = 3    N-1 = 444888

N-1 consists of 2 parts.   44...4  m times ;  88..8  (m) times.

1st part
10^m * 4 * [ 111...1  m times ]  =  10^m * 4 * (10^m -1) / (10 -1)
= 4/9  * 10^m * (10^m - 1)         as we apply sum of GP formula

2nd part =
8 * [ 1111..  m times] = 8 ( 10^m - 1) / (10 -1)
= 8/9  * (10^m - 1)

N - 1 = 4/9 * (10^m - 1) [ 10^m + 2 ]

N = 1 + 4/9 * (10^2m + 2 * 10^m - 10^m - 2)
N = 1 + 4/9 * 10^2m + 4/9 * 10^m  - 8/9
N   = 4/9 * 10^2m + 4/9 * 10^m  + 1/9

N = 1/3² * (2 * 10^m + 1)²

So N is a square of  A = (2 * 10^m + 1) /3

We have to prove that A is an integer.

A = [ 2 * (10^m - 1) + 3 ]  /3  =  1 + 2 * (10^m - 1)/3

We know that   10^m - 1 is like 9, 99, 999, 9999.... m times.
So 10^m - 1 is divisible by 9 and by 3.

A = an integer.
So N is q square of an integer = 1 + 2 * (10^m - 1)/ 3

m = 1,   sq root is 1 + 2 * 3
m = 2    s q root is  1 + 2 * 33
m = 3                    1 + 2 * 333 ....

thanx n u r welcom