Eleven Power

From the Pascal triangle (see given image), $(x + 1)^5 = x^5 + 5x^4 + 10x^3 + 10x^2 + 10x + 1$ With $x = 10$ , $(10 + 1)^5 = (10)^5 + 5(10)^4 + 10(10)^3 + 10(10)^2 + 5(10) + 1 = \boxed{161051}$ .

Notice that in the so-formed triangle,the set of numbers in the $n^{th}$ row represents a number sequence,that can be used to generate $11^{n-1}$ .[Like the 4th row has the sequence:1 3 3 1 => 1331= $11^3$ ]

Also,the set of numbers of the $n^{th}$ row are obtained from the $(n-1)^{th}$ row,by a method similar when we add 2 numbers:

[NOTE:this is an algorithm to compute the value of $11^n$ from the pascal's triangle.]

Start from the right.

Put a 1 at the extreme right end.

The $n^{th}$ row has n numbers,so,the (n-1)th row has (n-1) numbers.

Add 1 to the 2nd number from the right of the (n-1)th row to give the 2nd number from the right of the nth row.

Now,add the 2nd and 3rd numbers from the right of the (n-1)th row numbers,to give the 3rd number from the right of the nth row numbers.

Again add the 3rd and 4th numbers from the right of the (n-1)th row numbers,to give the 4th number from the right of the nth row numbers.

Like wise continue,and in the end,when there is no number left in the (n-1)th row to add with,put a 1 at the left most place of the nth row numbers.

$\textbf{But}$ ,here comes the difference:if in the process,we add 2 numbers from the (n-1)th row and get a 2 digit number ,then we only put the units digit at the place of the 2 digit number,and carry up the tens digit to the next addition.

Like wise we do....

By that way,we get the sequence-row for $11^5$ as 1 6 1 0 5 1,so $11^5=161051$

$\textbf{For those who want demonstration:}$

$11^4$ correspond to the sequence of digits 1 4 6 4 1.

So,in the extreme right is 1,as mentioned above in the algorithm.(for $11^5$ )

At the second place from the right is 1+4=5.

At the 3rd place from the right is 4+6=10,a 2 digit number.So,in order to compute $11^5$ ,we put the 0(of 10) at the place of 10,that is,3rd from right,and carry up the 1(of the 10) for the next addition(that is,6+4,by symmetry).

So the number,4th from the right will be 6+4+(carry over 1)=11,again a 2 digit number,so we put 1 there and again carry the 1 to the next addition.

Next is 4+1+(carry over 1)=6

And in the last,we now are not in a state of adding 2 nos,as only 1 number is left,that is,1. So,by algorithm,we put a 1 at the left most end of the sequence so generated.

So the final sequence becomes: 1 6 1 0 5 1,as wanted.

[Michael Huang's solution is good enough to qualify for a mathematical solution,but I felt that Chung Kevin deliberately included the Pascal's triangle so that we notice a pattern to calculate the powers of 11,so I wrote this solution.Also,this solution is dedicated to $\textbf{Indian Vedic Mathematicians}$ ,who started the $\textit{culture and tradition of observing number patterns and forming algorithms}$ .]