1. Jun 16, 2015

### vin300

I have recently discovered some mathematical rules by trial and error. Some of these may be already known to people of mathematics.
1. This one is common: Sum of all integers 1+2+3+...=0.5(n^2 + n)
2. Sum of odd integers: 1+3+5+7+9......= n^2
3. Sum of even integers: 2+4+6+8.....= n^2 + n eg: Sum of first two: 2+4= 2^2 +2 =6
4. Sum of gap odd integers(odd integers with one gap) : 1+5+9+13+17....= (n+1)(n+2) eg. 1+5=(1+1)(1+2)=6
5.a. Sum of gap even integers: 2+6+10+14....= 2n^2 eg 2+6+10= 2(3)^2=18
5.b. Sum of gap even integers: 4+8+12+16...= 4(1+2+3+4+...) eg:4+8+12= 4(1+2+3)= 24

2. Jun 16, 2015

### PeroK

3. Jun 16, 2015

### phinds

The algebraic derivations of all these is trivial, so yes, they are known.

4. Jun 16, 2015

### Stephen Tashi

If you want to develop such rules in a systematic fashion, try starting with a function like F(n) = n^2 and forming the series give by its differences: (F(1) - F(0)) + ( F(2) - F(1)) + (F(3) - F(2)) + ....(F(n+1) - F(n)). This series sums to F(n+1) - F(0).

For functions that are simple to write down, the "answer" F(n+1) - F(0) looks simple while the series whose terms are T(i) = F(i+1) - F(i) can look complicated.

5. Jun 17, 2015

### vin300

Sum of odd integers with one gap = n(2n-1)

6. Jun 17, 2015

### micromass

Staff Emeritus
Try to find a general formula now which includes all those cases:

$$a+ (a+b) + (a + 2b) + (a + 3b) + ... + (a + nb)$$

And try to find the sum of squares:

$$1 + 4 + 9 + ... + n^2$$

Or if you're up for a challenge, the more general

$$1 + 2^k + 3^k + ... + n^k$$