Question
(`2^2` + `4^2` + `6^2` +...........+ `20^2` ) = ?
Answer: Option C
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( `2^2` + `4^2` + `6^2` + .................+ `20^2`) = `(1 xx 2)^2` +`(2 xx 2)^2` + `(2 xx 3)^2` +..............+`(2 xx 10)^2`
= (`2^2` x `1^2`) + (`2^2` x `2^2`) + (`2^2` x `3^2`) + ........`( 2 xx 10)^2`
= `2^2` x [ `1^2` + `2^2` + `3^2` +...... + `10^2`]
= ` ( 4 xx1/6 xx 10 xx11 xx 21)` [`because (1^2 `+`2^2` + `3^2` +....... + `n^2`) = `1/6n(n+1)(2n +1)]`
= ( 4 x 5 x 77) 1540.
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