Goodness! This is giving me a headache...?

Question

Find the nth term of the following sequence:
1, 2, 4, 7, 11, 16, 22, 29..
I know that the differences are 1, 2, 3, 4, 5, 6, 7 and the differences of differences are 1 but I can't seem to find the nth term!

Errr... I didn't say to find the next number or find the differences. I mean the nth term. The formula for the sequence that can give you the number by it's position in the pattern.

You don't now what an nth term is? This is basic 7th grade mathematics!

Answer 1

n(n+1)/2 + 1

or n multiplied by (n plus 2), divided by 2, add 1

Answer 2

some people have put ((n(n + 1))/2) + 1

but if you put in 2, you should get 2, but instead you get 4

This is the correct formula, and i have checked it, and it will work

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

((1(1 - 1))/2) + 1 = ((1(0))/2) + 1 = (0/2) + 1 = 0 + 1 = 1
((2(2 - 1))/2) + 1 = ((2(1))/2) + 1 = (2/2) + 1 = 1 + 1 = 2
((3(3 - 1))/2) + 1 = ((3(2))/2) + 1 = (6/2) + 1 = 3 + 1 = 4
((4(4 - 1))/2) + 1 = ((4(3))/2) + 1 = (12/2) + 1 = 6 + 1 = 7
((5(5 - 1))/2) + 1 = ((5(4))/2) + 1 = (20/2) + 1 = 10 + 1 = 11
((6(6 - 1))/2) + 1 = ((6(5))/2) + 1 = (30/2) + 1 = 15 + 1 = 16
((7(7 - 1))/2) + 1 = ((7(6))/2) + 1 = (42/2) + 1 = 21 + 1 = 22
((8(8 - 1))/2) + 1 = ((8(7))/2) + 1 = (56/2) + 1 = 28 + 1 = 29

The next number would be

((9(9 - 1))/2) + 1 = ((9(8))/2) + 1 = (72/2) + 1 = 36 + 1 = 37

Answer 3

nth=8. 29+8=37

Answer 4

Xn = Xn-1 +(n-1)
When n = 2
Seq Xn-1 => X1 = 1, X2 = 2
When n=3
Seq Xn-1 => X2=2 --> X3 = 4

Therefore in words. The Nth value = nth to the minus 1 value plus n minus 1.

Answer 5

If we look at second differences, we find that they are constant. [i.e 1,2,3....]
This means that we can fit a quadratic to the points. f(x)=ax^2 +bx +c. Set f(0)=1, f(1)=2, f(2)=4 and solve for a,b,c
we get c=1,a=0.5, b=0.5
as we want the nth term substitute "(n)" in it
so we get it as.. 0.5 *(n)^2 + 0.5 *(n) + 1 where n starts from 0,1,2....

for ex.. if we want 4th term (7)
we put n = 3 and get the answer

Answer 6

37

Answer 7

Number = n
Subject Number = n1
Object Number = n2
Where, for example N1 = 1 and n2 = 2 to start
n1 = n2 + 1<1

Answer 8

37

Answer 9

nth term =
[n(n+1)]/2 + 1
i.e. work out n(n+1) then divide all by 2 then finally add 1.
I checked it and it works for all those I checked,

Answer 10

(n^2 - n + 2) / 2