Pattern

Determine the pattern to prove the identity
1 = 1
3+5 = 8
7+9+11 = 27

According to the given pattern in question
the general pattern is
(n^2-n+1) + (n^2-n+3) + (n^2-n+5).....(n^2-n+(2n-1)) = n^3

Here we need tp proof LHS = RHS
left hand side has n summands
n*n^2 - n*n + (1+3+5+.....2)
i.e n mumbers of n^2+1 and sum of odd numners only
Now, 1+3+5+.....(2n-1) equals to
(2*1-1)+(2*2-1)+(2*3-1)+....(2n+1)
2(1+2+3+4+....n)-1(1+1+1+1+1...)
2(n^2-n)/2 - n = n^2
n*n^2-n*n + n^2
n^3 = RHS

Pattern
1 = 1
2+3+4 = 1+8
5+6+7+8+9 = 8 + 27

1^3 = 1^3
(1^2+1) + (1^2+2) + 2^2 = 1^3 + 2^3
(2^2+1) + (2^2+2) + (2^2+3) + (2^2+4) + 3^2 = 2^3 + 3^3
((n-1)^2+1) + ((n-1)^2+2) + ((n-1)^2+3) + ((n-1)^2+4)+n^2 = (n-1)^3 + n^3
sum of integer upto n^2 - sum of integer upto (n-1)^2
(1+2+3+.....9)-(1.2...4) = (5+6+7+8+9)
[n^2*(n^2+1)]/2 =[(n-1)^2(n+1)^2+1]/2


Pattern
1=1
1-4= -(1+2)
1-4+9 = +(1+2+3)
1-4+9-16 = -(1+2+3+4)

general pattern is


1^2 - 2^2 + 3^2 - 4^2 + h^2 .......[(-1)^(n+1) - n^2] = (-1)^(n+1)(1+2+3...n)
For even
(1^2-2^2) + (3^2-4^2)+......[(n-1)^2 - n^2]
(1-2)(1+2) + (3-4)(3+4)......-1(n-1)+n
-1(1+2+3+4+.....(n-1)+n)

For odd
(1^2-2^2) + (3^2-4^2)+......[(n-2)^2-(n-1^2)] + n^2
(1^2-2^2) + (3^2-4^2)+......(n-2-n+1)(n-2+n-1) + n^2
-1(1+2)-1(3+4)+ -1[(n-2)+(n-1)^2] + n^2
-1{(1+2+3+4...(n-2)+(n-1)} + n^2
-n(n+1)/2 + n(n+1)
n(n+1)/2
1+2+3+4...n