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We can write the square of \[1`s\] and then find the square of this number.

Now we can construct and then solve by it, using this pattern method

$

{1^2} = 1 \\

{11^2} = 121 \\

{111^2} = 12321 \\

{1111^2} = 1234321 \\

{11111^2} = 123454321 \\

{111111^2} = 12345654321 \\

$

First we discuss about that the one square is one, eleven square is \[\;121\], one hundred eleven square is \[12321\],

Here we notice that it follows one pattern,

Suppose we take ${11^2}$ as equal to \[121\] , the answer should start with one and end with one.

We can add \[\left( {1 + 1 = 2} \right)\] that is in the middle position.

Also we can take ${111^2}$ is equal to \[12321\], answer should start and end with one and the middle term is \[\left( {1 + 1 + 1} \right)\] that is equal to \[3\] , and from second position of left and right is \[\left( {1 + 1 = 2} \right)\].

Also we can take ${1111^2}$ is equal to \[1234321\] , answer should start and end with one and the middle term is \[\;\left( {1 + 1 + 1 + 1} \right)\] that is equal to \[4\], and from third position of left and right this is of the form \[\left( {1 + 1 + 1 = 3} \right)\] and from second portion of left and right this is of the form \[\left( {1 + 1 = 2} \right)\]

Also we can take ${11111^2}$ is equal to \[\;123454321\], answer should start and end with one and the middle term is \[\left( {1 + 1 + 1 + 1 + 1} \right)\] that is equal to\[5\], and from fourth position of left and right this is of the form \[\left( {1 + 1 + 1 + 1 = 4} \right)\] and from third position of left and right this is of the form \[\left( {1 + 1 + 1 = 3} \right)\] and from second portion of left and right this is of the form \[\left( {1 + 1 = 2} \right)\]

Similarly we can find up to any \[1`s\] number.

Hence ${111111^2} = 12345654321$.