What is the recursive formula for the sum of a series of consecutive integers such as 1 + 2 + 3 + 4 + .....?

Answer 1

The value of 1 + 2 + ... + n = 0.5*n*(n+1)

Answer 2

The following formula will solve your integer problem.

S = n*(a0 + an) / 2,
where a0 = the first number in the sequence
an = the last number in the sequence
n = the number of terms in the sequence

For example, the sum of all integers from 1 to 100 is
S = 100 * (1 + 100) / 2 = 5050

Applying the formula to your example of 1 + 2 + 3 + 4
S = 4 * (1 + 4) / 2 = 10.................( 4 * 5 = 20 / 2 = 10)

Answer 3

Define two integers, X and Y such that Y = X + 1 Then, 2X + 4Y = 46 Substituting: 2X + 4(X + 1) = 46 2X + 4X + 4 = 46 6X + 4 - 4 = 46 - 4 6X = 42 X = 42/6 = 7 Y = X + 1 = 7 + 1 = 8 The integers are 7 and 8.

Answer 4

e.g. 1 + 2 + 3 + 4 + 5 + ...... + 98 + 99 + 100 = ?

1 + 2 + 3 + 4 + 5 + ...... + 98 + 99 + 100
= *(1 + 100) x 100 / 2
= 101 x 100 / 2
= 10100 / 2
= 5050

*(first item + last item) x total item / 2

Answer 5

It would be represented with the "sigma" summation symbol, and would look something like:
. 1
. Σ . . . n
n->∞
(the periods are for spacing only.)

The above example would start at 1 and add each consecutive integer, to infinity.

This example, of course, has been "solved" by Wesley in his post. For any given "n" in this summation, the sum up to and including n is:

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

Answer 6

If you are very specific about the formula being recursive , then it would be f(n) = n + f(n - 1) where f(1) = 1.

Answer 7

223xy X 3 to the integer of 15 to the 22 power...