JEE Math Practice Question
The value of $$\frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots .+100^2 \times 101}$$ is
- A.$$\frac{305}{301}$$
- B.$$\frac{306}{305}$$
- C.$$\frac{32}{31}$$
- D.$$\frac{31}{30}$$
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