Hexagonal Numbers

I've added to each hexagon six 'petals'. As we can see, for the $n^{th}$ hexagon, these petals grow just like triangular numbers of size $n-1$ . We know the formula for the $n^{th}$ triangular number is $\frac{1}{2}n(n+1)$ , so we substitute $n$ for $(n-1)$ to get $\frac{1}{2}n(n-1)$ . We have 6 of these petals plus the one central dot, which gives $1+3n(n-1)$ . Putting $n=20$ into this formula yields ${\color{#20A900}{\boxed{1141}}}$

We can deduce that each two consecutive hexagonal number has a difference which follows an arithmetic sequence - +6,+12,+18......

Which is apparently an AP with a first term 6 ,general difference(difference of difference)6,and term number 19.

So we can directly get the solution :1+1140(By using AP sum formula)=1141

In my case, I notice that I can use AP (Arithmetic Progression) for this question. As you can see, in every increase in number, the hexagon increase 1 outer layer. And since it is an hexagon, so it is an increment of 6. From here I could generate an AP formula.

According to AP formula, an=a1+(n-1)d. Sn=n(a1+an)/2. I let a1 = 6 which is the first increment of hexagon. d = 6, which is increment of every hexagon. and n = 19 for the 20th hexagon.

First I need to find how many dots are in the 20th hexagon. So a19=6+(19-1)6 and I get 114. And after I have the a19 value, I find the sum. S19=19(6+114)/2 which is equal to 1140. Never forgot the 1 miserable dot in the center.

And that's how I got the answer.

i found that $n$ th hexagonal no. was of the form $n^3-(n-1)^3$ .

$20^3-19^3 = 1141$

In first figure , there is only $1$ dot.
In second figure, there are $2+3+2$ dots.
In third figure, there are $3+4+5+4+3$ dots.
In fourth figure, there are $4+5+6+7+6+5+4$ dots.

So, we see that in each case the number of dots in the $n^{th}$ hexagon, is
$n+[n+1]+[n+2]+\cdots + [2n-1]+[2n-2]+\cdots+[n+1]+n$ dots.

So, in the $20^{th}$ hexagon, the number of dots are $20+21+\cdots+39+38+\cdots+21+20$ dots.
= $[[2×{\Large{\frac {39×40}{2}}}] -39] - [2×{\Large[\frac {19×20}{2}]}]$
= $[(39×40)-39] - [19×20]$
= $[1560 - 39] - [380]$
= $1521-380$
= $1141$

The median line which divides the hexagon in two halfs follows an aritmetic progression (1,3,5,7...). Therefore, the 20º term can be calculated as follows: an=a1+(n-1).r--> a20=1+19.2=39. That median line acts like a mirror, thereby the same amount of dots are found above and below it. Whatever direction you look from this parameter, you must found another aritmetic progression (38, 37, 36, .... , 20). The sum of its elements is: S=(a1+an).n/2---> S=(20+38).19/2=58.19/2=29.19=551. Since we have two of this A.P.s: 2.551=1102. Now, one just must add the amount of dots of the median line (39): 1102+39=1141.