Observe the following pattern
12 = 1/6 (1×(1+1)×(2×1+1))
12+22 = 1/6 (2×(2+1)×(2×2+1)))
12+22+32 = 1/6 (3×(3+1)×(2×3+1)))
12+22+32+42 = 1/6 (4×(4+1)×(2×4+1)))
And find the values of each of the following:
(i) 12+22+32+42+…+102
(ii) 52+62+72+82+92+102+112+122