What is the answer to this math problem?

Question

I have an Idea of what the answer may be, but I want to see if I'm correct.

ok.. well the questions is.
If you have the numbers (1,2) (3,4,5) (6,7,8,9) (10,11,12,13,14) and you keep following the pattern, what will the last number be if repeated to the 100th time??

Answer 1

You could just look at the pattern of the last number:
2, 5, 9, 14

Group 1 ends with 2
Group 2 ends with 5
Group 3 ends with 9
Group 4 ends with 14

The formula is n(n+3) / 2

The last number in the 100th group will be:
100(103) / 2 = 10300 / 2 = 5,150

Or you could look at the pattern of the first number:
1, 3, 6, 10, 15...

The well-known formula for this is:
n(n + 1) / 2

So if you find the 101st number and subtract 1 you should get the same answer:
101 (102) / 2 = 101 * 51 = 5151
Subtract 1 and you get:
5,150

Or you could take the answer for the 100th number and add 100:
100 (101) / 2 = 50 * 101 = 5,050
Add 100 and you get:
5,150

There you go 3 methods, and all the same answer.

Answer 2

Using the difference method for the first termof each group, I get it to be (n^2 + n)/2
So the first term of 100th sequence is (100^2 + 100)/2 = 5050
Each term has n+1 numbers in it
So 100th sequence will have 101 terms
So the list should be
5050, 5051, .......,5150

Answer 3

group 1 has to numbers
group 2 has 3 numbers
group 3 has 4 numbers
group 4 has 5 numbers
get the patern??
if you kept on group 5 would have 6 numbers
um lets see if this helps...
if you were small you get 2 cookies
you grow and get 3
you grow and get 4
you grow more and get 5
if you grew 100 times how many cookies would you get?? in this case you would have to grow 95 times more subtracting the ones i showed
i hope you understand
and good luck

Answer 4

2 + 3 + 4 + ......... + 100 + 101 = 101*102/2 -- 1 = 5150
the last number will be 100th
= ( 5050,5051, 5052, ........, 5149,5150).

Answer 5

it incredibly is a simultaneous equation concern. enable x: person, y: toddlers x+y=a hundred (a million) 50x+30y=3180 (2) ought to remedy the above 2 equations. Rearranging the 1st one: x=a hundred-y (3) changing equation (3) in (2): 50(a hundred-y)+30y=3180 (4) Simplifying (4) supplies: 5000-50y+30y=3180 and it incredibly is comparable to -20y=-1820 y=-1820/-20=ninety one changing y in (3) supplies x=a hundred-ninety one=9 consequently, no. of person is 9 and no. of toddlers is ninety one.

Answer 6

look at the end number of each set
n .... end number
1.........2
2........5
3........9
4.......14
you can write end number=1/2n^2+3/2n
end number=1/2(100^2)+3/2(100)=5000+150=5150

Answer 7

100+99+98+97+... just keep going

Answer 8

sum(i=1,100)= 101 * 50 = 5050

so resulting sequence would be (5050, 5051,...,5150)