Quant Question > If(12+22+32+…..+102)=385,then the valu...

If(1^{2}+2^{2}+3^{2}+…..+10^{2})=385,then the value of (2^{2}+4^{2}+6^{2} + …+20^{2}) is :

- a)770
- b)1155
- c)1540
- d)(385*385)

Correct answer is option 'C'. Can you explain this answer?

Sandhya Grewal
Jun 13, 2018 |

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If(12+22+32+…..+102)=385,then the value of (22+42+62+ …+202) is :a)770b)1155c)1540d)(385*385)Correct answer is option 'C'. Can you explain this answer?

(1^2+2^2+3^2+....10^2)=385

(2^2+4^2+6^2+....+20^2)=2^2(1^2+2^2+ 3^2+....+10^2 )

=4(385)

=1540

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(1^2+2^2+3^2+....10^2)=385(2^2+4^2+6^2+....+20^2)=2^2(1^2+2^2+ 3^2+....+10^2 )=4(385)=1540