k positive integers

Find the sum of k positive integers
1+2+3+4+........+k = s(k)
1+2+3+4+(k-1)+k+(k+1) = s(k+1)

using linear equation
(k+1)^2 = K^2 + 2K + 1
(k+1)^2 - k^2 = 2K + 1-----------1

When k = 1
2^2-1^2 = 2*1 +1
When k = 2
3^2-2^2 = 2*2 +1
When k = 3
4^2-3^2 = 2*3 +1
When k = 4
5^2-4^2 = 2*4 +1
When k = k
(k+1)^2 - k^2 = 2K + 1

Now adding both sides
(2^2-1^2)+(3^2-2^2)+.........(k^2-(k-1)^2) + (k+1)^2 - k^2 = 2(1+2+3+4....k) + (1+1+1+1+ to k)


(k+1)^2 - 1^2 = 2s(k) + k
K^2 + 2k +1 -1 = 2s(k) + k
s(k) = (K^2 + k)/2 - k(k+1)/2

Find the sum of k^2 positive integers
1^2 + 2^2 + 3^2 + 4^2 +.......k^2 = s(k^2)
1^2 + 2^2 + 3^2 + 4^2 + (k-1)^2 k^2 + (k+1)^2 = s((k+1)^2)

using equation.
k^3 - (k-1)^3 = 3k^2 - 3K + 1

put k = 1,1,2,3 ....... respectively
0^3 - (0-1)^3 = 3*0^2 - 3*0 + 1
1^3 - (1-1)^3 = 3*1^2 - 3*1 + 1
2^3 - (2-1)^3 = 3*2^2 - 3*2 + 1
..
..
..
k^3 - (k-1)^3 = 3k^2 - 3*k + 1
-------------------------------Adding both sides
k^3 = 3(0^21^2 + 2^2 + 3^2 + 4^2 + k^2) - 3(1+2+3.....k) + k
k^3 = 3s(k) - 3k(k+1)/2 + k
k^3 + 3k(k+1)/2 - k = 3s(k)
s(k) = K^3/3 + (3k^2 + 3k)/6 - k/3
(2k^3 + 3k^2 + 3K - 2k)/6
.....
.....
s(k) = (k(2K+1)(k+1))/6


Claculate the sum of first k odd positive integer
1+3+5+7+......(2k-1)
First adding all numbers from 1 + (2k-1) and subtracting even numbers
1+2+3+4+.....(2k-1)-[2+4+6+....2k-2]
((2k-1)^2 +(2k-1))/2 - 2[(k-1)^2)+(k-1)]/2
[(2k-1).2k]/2 - 2/2(k-1) * k
2k^2 - k^2
k^2