Question

Find the first triangular number which is greater than 1000????

Answer 1

1,3,7,13 ... those things right?

Answer 2

1,3,6,10

Answer 3

yeah, 1,3,6,10 :)

Answer 4

pascals triangle might be useful

Answer 5

I know it is be some sort of inequality.

Answer 6

I don't know what pascals triangle is.

Answer 7

Its all the way at the back of my text.

Answer 8

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

the 3rd inset is our trinagle numbers

nC2

Answer 9

n(n-1)
------ > 1000
2

n^2 -n > 2000

Answer 10

n^2 -n -2000 > 0

Answer 11

that equals 0 when n=45.xxx; so 46 is my bet

Answer 12

45 my text says.

Answer 13

45C2 = 45(22) = 990

Answer 14

the triangles start with 2C2 so yes; 45 is it :/

Answer 15

another method i recall has to do with finding the degree of differences

1 3 6 10 15
2 3 4 5
1 1 1

and the first term in each row is the coeff of a factorial exapnasion for lack of a more descriptive term

1 + 2n + 1n(n-1)

1 + 2n + n^2 -n

n^2 +n + 1 but im sure im remembering it a little off tho

Answer 16

>>> i = 1
>>> n = i
>>> while n < 1000:
... i += 1
... n += i
...
>>> i
45

Answer 17

Python

Answer 18

\[t_n=t_{n-1}+n;\ t_0=0\]
\[t_n=t_{n-2}+(n-1)+n\]
\[t_n=t_{n-3}+(n-2)+(n-1)+n\]
\[t_n=t_{n-4}+(n-3)+(n-2)+(n-1)+n\]
\[t_n=t_{n-r}+(n-(r-1))+(n-(r-2))+(n-(r-3))+...+(n-3)+(n-2)+(n-1)+n\]
when r=n
\[t_n=t_{0}+(1+2+3+...+(n-3)+(n-2)+(n-1)+n\]
\[t_n=\frac{n(n+1)}{2}\]
\[\frac{n(n+1)}{2}>1000\]
\[n(n+1)>2000\]
\[n^2+n-2000>0\]
\[n>\frac{-1+\sqrt{8001}}{2}::\frac{88.5}{2}\to\ 44.25\]

Answer 19

.... the antiPython lol

Answer 20

Yes, in some ways in makes me feel dirty. But it's fast and easy.