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Question: 11<sup>2</sup> + 12<sup>2</sup> + 13<sup>2</sup> +…..20<sup>2</sup> is equal to...

112 + 122 + 132 +…..202 is equal to

A

2481

B

2487

C

2485

D

2483

Answer

2485

Explanation

Solution

112 + 122 + 132 +…..+ 202

= (12 + 22 + 33 +……+ 202) – (12 + 22 + 32 +…..+ 102 )

= – [n(n+1)(2n+1)6]n=10\left[ \frac { \mathrm { n } ( \mathrm { n } + 1 ) ( 2 \mathrm { n } + 1 ) } { 6 } \right] _ { \mathrm { n } = 10 }