(112 + 122 + 132 + … + 202) = ?

Question: (112 + 122 + 132 + … + 202) = ?
[A].

385

[B].

2485

[C].

2870

[D].

3255

Answer: Option B

Explanation:

(112 + 122 + 132 + … + 202) = (12 + 22 + 32 + … + 202) – (12 + 22 + 32 + … + 102)

	Ref: (12 + 22 + 32 + … + n2) =	1	n(n + 1)(2n + 1)
		6

=		20 x 21 x 41	–	10 x 11 x 21
		6		6

= (2870 – 385)

= 2485.