Simple logic pattern

Notice that at the end of $n^{th}$ row, the number will be $n^2$ , therefore at the row 100, the last number will be 10000, and that row start from where row 99 left off at 9801 (i.e. row 100 starts with 9802 and ends with 10000). Since that row contains 199 number, the number in the middle is then the number at position of 100 of row 100, which means the middle term is $9802+99 = 9901$

We note that the middle terms of the first four rows are $1,3,7,13$ . This implies that the middle term $a_n$ is given by $a_1 = 1$ and $a_n = a_{n-1}+2(n-1)$ for $n > 1$ . Therefore,

$\begin{aligned} a_{100} & = 1 + 2 + 4 + 6 + ...+ 2(99) \\ & = 1 + 2(1+2+3+...+99) \\ & = 1 + 2 \times \frac{99(100)}{2} \\ & = \boxed{9901} \end{aligned}$

n² - (n - 1)

Number in the middle of the row is going to be:

n(n-1)+1

so 100th row is 100(100-1)+1 = 9901

The pattern can be represent as $f(x)=x^2-x+1$

So $f(100)=100^2-100+1$ $=10000-100+1$ $=9901$

All we have to do is find the sum of first 99 even natural numbers and add 1 to the result.
= $\left(\dfrac{2 + 198}{2} × 99\right) + 1 = \boxed{9901}$ .

The middle term of the $n^{\text {th}}$ row is $\displaystyle\left\lceil\frac{n^2+(n-1)^2}{2}\right\rceil$ . So the answer is $\displaystyle\left\lceil\frac{100^2+99^2}{2}\right\rceil=9901$ .

M(n) = n(n - 1) + 1

M(100) = 100(99) + 1
= 9900 + 1
= 9901

the numbers in each row is given by the connection a(n)=1+2(n-1) summing up the numbers in each row will give us the last number to the n' row. so s(100)= (200)×100/2=10000. at the 100th row there are 199 numbers. so the middle number would be the 100th in the row: 1000-99=9901

Last Number on nth-row = n^2

Number of numbers between middle & last = n-1

middle number = (n^2)-(n-1)

n=100 ==> 10,000-99 = 9901

Let x be the row number. We notice that the end of each line = ${x}^{2}$ . Because of the way the numbers are set up, you can logically make the equation ${x}^{2}-(x-1)$ . For 100, this is 10,000-99 which is 9901.