Question

Find the value of `11^2+12^2+13^2++20^2dot`

Answer 1

Correct Answer - 2485
We know that `(1^(2)+2^(2)+3^(2)+ ...+n^(2))=(1)/(6)n(n+1)(2n+1).`
`therefore " given sum"={1^(2)+2^(2)+...+(10)^(2)+(11)^(2)+...+(20)^(2)}-{1^(2)+2^(2)+..+(10)^(2)}`
`=((1)/(6)xx20xx21xx41)-((1)/(6) xx 10xx11 xx21)=(2870-385)=2485.`