Square root of what did you just say?

Can anyone give a proper solution? I found it by pattern.

Pretty easy actually. 49=7^2 4489=67^2 444889=667^2 hence we see, for the number of 8's there are same number of 6's in the sq root. Hence there are 2014 6's and one 7 at the end of the sq root of the given term their sum=2014*6+7=12091

There is nothing special here about 2014 or 2015, mathematically speaking, so we see the general case. After playing around a bit, we notice a pattern given below.

We use induction to prove that $f(n)={\underbrace{66\ldots6}_{n-1}7}^2=\underbrace{44\ldots4}_{n}\underbrace{88\ldots8}_{n-1}9; n\in\mathbb{N}$

As usual, we prove the statement to be true for $n=1$ and assume it to

be true for $n=2, 3, \ldots , k$ , and then prove it for $n=k+1$ .

Notice that $f(3)-f(2)=444889-4489=440400$ , and also that $667^2-67^2=(667-67)(667+67)=(600)(734)=440400$ . So if we can show that $f(k+1)-f(k)=\underbrace{44\ldots4}_{k+1}\underbrace{88\ldots8}_{k}9-\underbrace{44\ldots4}_{k}\underbrace{88\ldots8}_{k-1}9$ , we're done.

$f(k+1)-f(k)= {\underbrace{66\ldots6}_{k}7}^2-{\underbrace{66\ldots6}_{k-1}7}^2 =(\underbrace{66\ldots6}_{k}7-\underbrace{66\ldots6}_{k-1}7)(\underbrace{66\ldots6}_{k}7+\underbrace{66\ldots6}_{k-1}7) \\ =(6\times10^k)(7\underbrace{33\ldots3}_{k-1}4) \\ =6\times7\underbrace{33\ldots3}_{k-1}4\times10^k \\ =44\underbrace{00\ldots0}_{k-1}4\underbrace{00\ldots0}_{k} \\ =\underbrace{44\ldots4}_{k+1}\underbrace{88\ldots8}_{k}9-\underbrace{44\ldots4}_{k}\underbrace{88\ldots8}_{k-1}9.$

QED.