Given that 1² + 2² + 3² + ... + n² =

Given that 1² + 2² + 3² + ... + n² = n(n + 1) (2n + 1) , then
6
10² + 11² + 12² + .... + 20² is equal to

Given in question ,
1² + 2² + 3² + ... + n²

=	n(n + 1)(2n + 1)
	6

∴ 10² + 11² + 12² + .... + 20² = (1² + 2² + 3² + .... + 20²) – (1² + 2² + 3² + ... + 9²)

Required answer =	20(20 + 1)(40 + 1)	-	9(9 + 1)(18 + 1)
	6		6

Required answer =	20 × 21 × 41	-	9 × 10 × 19
	6		6

∴ Required answer = 2870 – 285 = 2585

Given that 1² + 2² + 3² + ... + n² =	n(n + 1) (2n + 1)	, then
	6