What is the general formula for the series: 1, 2, 4, 7, 11, .....?

Question

What is the general formula for the series: 1, 2, 4, 7, 11,....??? 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....pls help me...

Answer 1

The pattern seems to be:

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

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[2] = a[1] + 1 = 1 + 1 = 2
a[3] = a[2] + (3 - 1) = 2 + 2 = 4
a[4] = a[3] + (4 - 1) = 4 + 3 = 7
a[5] = a[4] + (5 - 1) = 7 + 4 = 11

And so it does, so the recursive definition is:

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

If you want a closed form, it's going to be a quadratic relation, as we want the points that go through (1,1), (2, 2), (3, 4).

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

f(1) = a(1)^2 + b(1) + c = 1
f(2) = a(2)^2 + b(2) + c = 2
f(3) = a(3)^2 + b(3) + c = 4

Three equations, three unknowns.
a + b + c = 1
4a + 2b + c = 2
9a + 3b + c = 4

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

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(1) = (1/2) (1 - 1 + 2) = (1/2)(2) = 1
f(2) = (1/2) (4 - 2 + 2) = (1/2)(4) = 2
f(3) = (1/2) (9 - 3 + 2) = (1/2)(8) = 4
f(4) = (1/2) (16 - 4 + 2) = (1/2)(14) = 7
f(5) = (1/2) (25 - 5 + 2) = (1/2) (22) = 11

It seems to satisfy the series!

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

Answer 2

1+1=2
+2=4
+3=7
+4=11
+5=16

Answer 3

Take the numbers 1, 2, 3, 4, 5, ... as the "adding series".
Have a look at your series and start with 1.
Take the first number of your "adding series" and add them up: 1+1 = 2. This is the second number of your series.
Take the second number of your "adding series" and add it to the result: 2 + 2 = 4.

Go on with this rule: 4 + 3 = 7.
And:7 + 4 = 11
and so forth...

Answer 4

1
1 + 1 = 2
2 + 2 = 4
4 + 3 = 7
7 + 4 = 11
you add the next higher integer to the last number to get the next number.

Answer 5

1+1=2
2+2=4
4+3=7
7+4=11
11+5=16.........n so on
the no. to be added keeps on increasing by 1

Answer 6

S = 1+2+4+7+11+..........T(n)
S = 1+2+4+7+..........T(n-1) + T(n)

Subtracting both we get,

0 = 1 + 1+2+3+4+........(n-1) - T(n)

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

T(n) = [n(n-1) + 2]/2

Answer 7

x subn+1 = x sub(n) +n where n=0,1,2,3..

so the first few are
1=1+0
2=1+1
4=2+2
7+4+3