the first spiral have length "2" the second spiral have length "4" the third spiral have length "6"

When you observe it you could see that the length will increase by 2 due to the spiral movement of the drawing. the first spiral have "2'' as the number at the bottom-right corner the second spiral have "12" as the number at the bottom-right corner the 3rd spiral have "30" as the number at the bottom-right corner

In as square to determine the area just simple square the side. at the diagram it is a square and has numbers at its side. so when you square the length of the side you will get the number at the bottom-left corner. so to know the number at the bottom-right corner, you need to know the number at the bottom-left corner and then subtract the length to know determine the number at the opposite side or the bottom-right corner which is like subtracting the distance. as the pattern continues the 15th length would be 30. then square the length then subtract the length. just like having a formula for "x" as the length. x squared minus x would be the number at the bottom-right corner. 30 squared minus 30 would give you 870.

If we take away every number from the spiral greater than or equal to the $n$ -th bottom-right corner number, we are left with a rectangle of size $2n(2n-1)$ . Since we started with $0$ , the highest number left must be $2n(2n-1) - 1$ , and therefore the $n$ -th bottom-right corner number must be the next in line, namely $2n(2n-1)$ . In the case of $15$ , this gives us $29\cdot 30 = 870$ , which is the answer for this problem.

It is clear to see that in order to get to 12 from 2, we need to plus 2 to get the bottom-left corner, plus another 2 to get the top-left corner, plus 3 to get the top-right corner and plus another 3 to get the bottom-right corner. It can be concluded as 2+(2+3) * 2=12 Similarly, the third bottom-right number is 12+(4+5) * 2=30 since the spiral expands, sub in the previous equation to get 2+(2+3+4+5) * 2=30. Therefore, the nth bottom-right number is 2+(2+3+........+2n-2+2n-1) * 2 The 15th bottom-right number is 2+(2+3+......+28+29) * 2= 2+31*28=870

looking at first numbers we see that this numbers are 2, 12, 30, 56, 90... and difference between them is 2, 2+8, 2+8 2, 2+8 3..... 2+8 k. Then number on position k is 2+ (2+8)+(2+8 2)+ (2+8 3)+....+(2+8 (k-1))=2 k+8 (0+1+2+3+...+(k-1))=2k+8 (k (k-1)/2)=2k+4k(k-1). Then number on 15th position is 2 15+4 15*14=870

The numbers at the bottom left are: $2 \rightarrow 12 \rightarrow 30 \rightarrow \ldots \rightarrow X$

The diference between consecutive numbers forms an arithmetic progression, whose general term is:

$2 - 0 \rightarrow 12 - 2 \rightarrow 30 - 12 \rightarrow \ldots$ = $2 \rightarrow 10 \rightarrow 18 \rightarrow 26 \rightarrow \ldots$

$An = A1 + (n-1)\times r = 2 + (n-1)\times 8$

So, the 2nd number can be write as: $\displaystyle \sum_{i=1}^2 Ai = 2 + 10 = 12$

The Nth: $\displaystyle \sum_{i=1}^n Ai$

The formula from sum of an arithmetic progression is:

$\frac{(A1 + An) \times n} {2}$ = $\frac{(A1 + (A1 + (n - 1) \times r)) \times n }{ 2}$

$X = \displaystyle \sum_{i=1}^{15} Ai = \frac{(2 + 2 + 14 \times 8) \times 15}{2} = \frac{(112 + 4) \times 15}{2} = 870$

So the 15th term is 870.

We can define in the same manner an upper-right corner . By simply watching the spiral, we see that those are the squares of odd numbers. This can be proven by mathematical induction easily.

So, the k th upper right corner is $\left(2k-1\right)^2$ , which in turn means that the k th lower-right corner is $\left(2k-1\right)^2 + \left( 2k-1 \right) = \left( 2k-1\right)\left( 2k\right).$

For $k=15$ we get the number is $29 \cdot 30= 870$ .

Let the $n^{th}$ bottom-right corner number are $U_n$ at the question, we know that $U_1 =2$ and $U_2 =12$ If we continue the spiral pattern until we get $U_3$ , we will get $U_3=30$ . Observe that $U_1 = 2 = 1 \times 2 =(2 \times 1 - 1) \times (2 \times 1 )$ $U_2 = 12 = 3 \times 4=(2 \times 2 - 1) \times (2 \times 2 )$ $U_3 = 30 = 5 \times 6 =(2 \times 3 - 1) \times (2 \times 3 )$

With this pattern, we deduce that $U_n = (2n - 1) \times (2n)$ for $n \geq 1$

Now, we can find $U_15$ Hence, $U_{15} = (2 \times 15 - 1) \times (2 \times 15 ) = 29 \times 30 = \boxed{870}$

Continuing the pattern will lead us to more bottom-right corners 2 - 12 - 30 - 56 - 90. It can be observed that the difference within the numbers are 10 - 18 - 26 - 34, whose shares a common difference of 8. Based on the observations, we can assume that the pattern satisfies $f(x)=ax^{2}+bx+c$ , where $a,b,c$ is a real number.

Put in the first, second, and third value of the series, which will get us to:

$a+b+c = 2$

$4a+2b+c=12$

$9a+3b+c=30$

By solving the equations, we get $a=4, b=-2, c=0$ , thus the n-th term of the pattern can be expressed as $4n^{2}-2n$ . Therefore, the 15th term is $4 \times 15 \times 15 - 2 \times 15 = 870$

The first $3$ bottom-right corners:

$2 = 2^{2} - 2$ $12 = 4^{2} - 4$ $30 = 6^{2} - 6$ And so on. By pattern recognition, we have that the $n$ th bottom-right corner is equal to $(2n)^{2} - 2n = 2n(2n - 1)$ . As we are looking for the $15$ th bottom-right corner, then we are looking for $2(15)(2(15) - 1) = \boxed{870}$ .

By conitinuing drawing the diagram it is found that the third term is 30. let, t {n} be the nth bottom right corner number. we have, t {1}=2, t {2}=12, t {3}=30 let, s {n} be the sum of first n bottom right corner number. Now, s {15}=2+12+30+.........................+t_{15}

-s {15}= -2-12+........................ -t {14} -t_{15}

0   =(2+10+18+..................upto 15 terms) - t_{15}

t {15}=2+10+18+.............upto 15 terms t {15}=870 [Simply using the addition formula of A.P.]

Therefore, the required fifteenth bottom right corner number of the given diagram is 870.

If one analyses the situation, he will observe that the the n-th bottom right corner number is actually equal to (2n)^{2} - (2n) ...Thus for n=15 it is 30^{2}-30 = \boxed{870}

first observe the top right numbers. They are the squares of odd numbers. Hence using the formula for A.P 15th top right number is 29 ( T = a+ (n-1) d => 1 +(15-1) 2 = 29). And every bottom right number is nth top right number + n. Which gives us the answer as 29*29 +29 = 870.

If we make one more spiral then we can observe that top - right - corner is as square of odd numbers.

So, for 15th top - right - corner = (29)^2 = 841.

And, quantity of numbers between top - right - corner and bottom - right - corner is increasing as = 2 * (n - 1) where n is the number of spirals.

Therefore, for 15th bottom - right - corner (say x) , we have to do the following process;

x = 841 + 2 * (15 - 1) +1

= 842 +28

= 870 (Ans)

The problem can be solved by noting that the generator of the $n^{th}$ bottom-left corner ( $G_L(n)$ ) is given by:

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ G_L(n) = (2n)^2 - 1$

Notice further that the $n^{th}$ bottom-right corner is $(2n -1 )$ less than the $G_L(n)$ . Therefore, the bottom-right corner is given by:

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ G_R(n) = G_L(n) - (2n -1)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ G_R(n) = (2n)^2 - 1 - (2n -1) = (2n)^2 - 2n = 2n(2n-1)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ G_R(n) = (2n)^2 - 2n = 2n(2n-1)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ G_R(n) = 2n(2n-1)$

Now, we simply need to substitute 15 to $G_R(n)$ . Thus we have,

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ G_R(15) = 2\times 15 \times (2\times 15-1)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ G_R(15) = 30 \times (29)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ $G_R(15) =$ 870

Let the k-th bottom-right corner number be $T_k$ By observing the pattern, we can understand that, $T_0 = 0$ and $T_{k+1} = T_k + 8k + 2$ Let, $U_k = T_k - T_{k-1} = 8(k-1)+2$ Then, $\sum_{i = 1}^{k} U_i = T_k - T_0 = T_k$ $= 8 \sum_{i = 1}^k (k-1) + 2 k$ $= 4k(k-1) + 2k = 2k(2k-1)$

So, $T_{15} = 2\times 15\times 29 = \boxed{870}$

We first define a number in the upper-right corner as a number with a rightward arrow pointing at it and a downward arrow pointing out of it. We also define a number in the bottom-left corner as a number with a leftward arrow pointing at it and an upward arrow pointing out of it.

By observation, the $n^{th}$ bottom-left corner number is given by $(2n)^{2}$ and the $n^{th}$ upper-right corner number is given by $(2n-1)^{2}$ . With respect to these two, the $n^{th}$ bottom-right number is given by $\frac{1}{2} \times ((2n-1)^{2} + (2n)^{2} - 1)$ . By substitution, the $15^{th}$ bottom-right number is $\frac{(2(15)-1)^{2} + (2(15))^{2} - 1)}{2} = \boxed{870}$

we can use this equeation to solve this problem : 2n(2n-1) if n=1, the value is 2, if n=2, the value us 12, if n=3, the value is 30, .... so if n=15, the value is 870.

Let us denote the number in the $n$ -th bottom-right corner by $a_{n}$ . First we are going to prove by Mathematical Induction the following statement: $\quad\quad\quad\quad\quad\quad\quad\quad\quad$ "For $n\ge 2$ the following recursion formula is always true $a_{n} =a_{n-1}+8n-6$ where $a_{n}$ is obtained by moving from $a_{n-1}$ to the left $2n-2$ units, then $2n-2$ up, $2n-1$ units to the right, and $2n-1$ units down." $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$ For $n=2$ the previous statement is obvious, because $a_{1} =2$ and $a_{2}=12.$ and you get the $a_{2}$ from $a_{1}$ by moving first 2 units left, 2 units up, 3 units to the right and 3 units down. Now let us assume that for an integer $k\ge 2$ the statement is true so $a_{k}=a_{k-1}+8k-6$ and $a_{k}$ is obtained from $a_{k-1}$ by moving $2k-2$ units right, $2k-2$ up, $2k-1$ right and $2k-1.$ down. Look at the figure below From that diagram is easy to realize that to move from $a_{k}$ to get to $a_{k+1}$ along the spiral one has to move first $2k$ units to the left, $2k$ units up, $2k+1$ to the right and $2k+1$ down, and, therefore, $a_{k+1}=a_{k}+8k+2=a_{k}+8(k+1)-6.$ So the recursion is true. $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$ Now, adding all the equalities of the form $a_{k}=a_{k-1}+8k-6$ side by side from $k=2$ to $k=n$ and cancelling terms that appear repeated in the left and the right side of the resulting equality, we obtain $a_{n}=a_{1}+8(2+3+4+...+n)-6(n-1),$ therefore $a_{n}=4n^{2}-2n.$ So the answer is $a_{15}=4(15)^{2}-2(15)=870.$

The series is 2x(2x-1)

So the answer is 30 x 29 = 870

(For next, we can use term BR as the bottom-right corner number and BL as the bottom-left corner number) I'm not using any theorems, I only read the pattern inside those numbers. 1) I first realize that the BL must be the square even number. The first of its kind is 4 = 2^2. And if you continue the spiral, you will see that the second is 16 = 4^2, the third is 36 = 6^2, and so on. So I think the 15th BL must be the square of the 15th even number, namely 30. Shortly, the 15th BL is 900. 2) I then observe the difference of BL and BR at the same stage of spiral. Consecutively, from the first stage of spiral, the pattern is: 4 - 2 = 2 = 1 \times 2, 16 - 12 = 4 = 2 \times 2, 36 - 30 = 6 = 3 \times 2. And so on. 3) I just guess that the difference of BL and BR must give an sequence, formulated by d = 2n, such that n is the number of stage at the spiral. 4) Conclution: the difference between the 15th BL and 15th BR must be 2 \times 15 = 30. And since the 15th BL is 900, thus the 15th BR is 900 - 30 = 870. 5) Unpredictably, it's correct. :)

just spread the spiral chord for a few more numbers.and you see the numbers proceed as 2,12,30,56....i.e. the sequence has common term t(n)=2 n (2n-1)....So you can easily calculate the 15th term to be 870...

bottom-right corner numbers are 0,2,12,30............... deference b/w them respectively are 2,10,18....... now, 2,10,18........ have a regular pattern that is 2 * 1=2, 2 * 5=10, 2 * 9=18, ...... ........ now , 1,5,9.......in A.P. so, addition of these first 15 number is- Sn=n/2((2a+(n-1)d) where- n=15, a=1, d=4 , So, S(15)=15/2((2 * 1 + (15-1) * 4) = 435 So, 15th bottom corner number is = S(15) * 2 =870 Ans.=870

By expanding the pattern, it builds a sequence $K = 2, 12, 30, 56, ...$ Noticing that: $2 = 2 \times 1 = 2 \times (1)$ $12 = 2 \times 6 = 2 \times (1+5)$ $30 = 2 \times 15 = 2 \times (1+5+9)$ $56 = 2 \times 28 = 2 \times (1+5+9+13)$ It's inferred that each n-th member of K is formed by twicing the sum of the terms of the AP $a_{n} = 1 + 4 \times (n - 1)$ Hence: $K_{n} = 2 \times S_{n}= 2 \times \frac{n}{2} \times (a_{1} + a_{n})$ $K_{n} = n \times [1 + [1 + 4 \times (n - 1)]]$ $K_{n} = 4n^2 - 2n.$ Therefore: $K_{15} = 4 \times 15^2 - 2 \times 15$ $K_{15} = \boxed{870}$

let n be the order of the spiral. (ie n=1 for the first spiral and n=2 for the second spiral) we are asked to find what is the bottom right corner number for the 15th spiral (ie n=15). let N be the number of elements in each spiral. thus it is seen that N=2n -> (1) for the 15th spiral order N=30; The number that lies at the bottom right corner follows a pattern : 2,12,30... it is a product of N*(N-1). Thus for N=30, the number is 870

f(i) = f(i-1) + (i-1)*8 + 2 with f(1) =2

a simple C program to ennumerate the numbers

include<stdio.h>

int main(){

int i;
int sum=2;
int count=10;
for(i=0;i<15;i++){
    sum+=count;
    count+=8; // only thing to note is that difference among corner numbers is increasing by 8
    printf("%d ",sum);
}

}

The 1st such number happens to be 2 which can be written as 1X 2, the next such number is 12 which again can be represented as 3 X 4. For developing a pattern we follow the brute force approach and find the next such number which is 30 which in turn again can be represented as 5X6. So the numbers called into question are essentially the product of successive numbers encountered in simple counting beginning with an odd number and ending with the even number. Without the loss of generality the first number 1 may be written as 2n-1 and 2 may be represented as 2n where n=1, and subsequent numbers may be obtained by simply putting the position of the number for example the second number is obtained by putting n=2, the numbers being 3 and 4 and their product 12. Thus the 15 such number is obtained by the product of 2n-1 and 2n with n=15, the answer being 870.