What is the general formula the series 1, 2, 4, 7, 11,....for n>=0?

Question

What is the general formula for the series: 1, 2, 4, 7, 11,....for n=0, 1, 2, 3, ....??? Could anyone help me with this problem, I've been trying to use trial-and-error but I cannot fine the general formula for this series...What I mean with the general formula is an expression in terms of n which satisfies this series which starts with n=0....pls help me...

Answer 1

The pattern seems to be:

a[0] = 1
a[n] = a[n - 1] + n

Note: the square brackets are used to represent a subscript. That is, a[7] would represent the 7th term in the sequence, and a[n] represents the n-th term.

Let's find out if this holds true for the first 5 terms.

a[1] = a[0] + 1 = 1 + 1 = 2
a[2] = a[1] + 2 = 2 + 2 = 4
a[3] = a[2] + 3 = 4 + 3 = 7
a[4] = a[3] + 4 = 7 + 4 = 11

And so it does, so the recursive definition is:

a[1] = 1
a[n] = a[n - 1] + n

If you want a closed form, it's going to be a quadratic relation, as we want the points that go through (0,1), (1, 2), (2, 4).

Let f(n) = an^2 + bn + c. Then

f(0) = a(0)^2 + b(0) + c = 1
f(1) = a(1)^2 + b(1) + c = 2
f(2) = a(2)^2 + b(2) + c = 4

Three equations, three unknowns.
c = 1
a + b + c = 2
4a + 2b + c = 4

I won't show you the details, but solving this system of equations yields the following solution.

a = 1/2
b = 1/2
c = 1

Which means f(n) = (1/2)n^2 + (1/2)n + 1

We can make this form cleaner by factoring and changing its form:

f(n) = (n^2)/2 + (n/2) + 2/2

So our final closed form is:

f(n) = (1/2) (n^2 + n + 2)

Let's test it out!
f(0) = (1/2) (0^2 + 0 + 2) = (1/2)(2) = 1
f(1) = (1/2) (1^2 + 1 + 2) = (1/2)(4) = 2
f(2) = (1/2) (2^2 + 2 + 2) = (1/2)(8) = 4
f(3) = (1/2) (3^2 + 3 + 2) = (1/2)(14) = 7
f(4) = (1/2) (4^2 + 4 + 2) = (1/2) (22) = 11

It seems to satisfy the series!

Answer 2

The sum of n natural numbers is given by

n (n+1) /2.

Denote the sum as Sn

S1 = 1, S2 = 3, S3 = 6, S4 = 10, S5= 15 -----Sn = n(n+1)/2

The series is 1, 3, 6, 10, 15.........n (n+1)/2

The given series is 1, 2, 4, 7, 11 etc.

The first term of the series is S1- 0

The second term of the series is S2 -1 and so on.

The nth term of the series is Sn - ( n-1)

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

This is simplified further into ½ { n^2 -n +2)

Check

The fifth term is ½(25 - 5 +2) = 11 and hence correct.

Answer 3

the formula is:
T(n)=T(n-1)+(n+1) ,T(0)=1
Where T(n) denotes the n th term
T(n-1) denotes the (n-1) th term
T(0) denotes the term for n=0

Answer 4

a(n)=a(n-1)+n
a(0)=1
but I am not sure if this is correct or if there should be no a in the second part

Answer 5

i dont think there can be a general formula for this series - it just doesnt exist.