Question 125

Given $$2^{2}+4^{2}+6^{2}+........+40^{2} = 11480$$, then the value of $$1^{2}+2^{2}+3^{2}+........+20^{2}$$ is:

Solution

Given : $$2^{2}+4^{2}+6^{2}+........+40^{2} = 11480$$

=> $$2^2[1+2^2+3^2+.....+20^2]=11480$$

=> $$1^{2}+2^{2}+3^{2}+........+20^{2}=\frac{11480}{4}$$

=> $$1^{2}+2^{2}+3^{2}+........+20^{2}=2870$$

=> Ans - (A)

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