A) 0.331
B) 0333
C) 0334
D) 0332
Correct Answer: D
Solution :
\[\frac{\sum\limits_{i=1}^{n}{{{i}^{2}}}}{\sum\limits_{i=1}^{n}{i}}=\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}=\frac{2n+1}{3}\] For \[n=1,2,3,....,1000\] Value of \[\frac{2n+1}{3}=\frac{3}{3},\frac{5}{3},\frac{7}{3},.........,\frac{2001}{3}\] respectively. Out of\[\frac{3}{3},\frac{5}{3},\frac{7}{3},.........,\frac{2001}{3}\] only first tern \[\left( \frac{3}{3}=1 \right),\] fourth term\[\left( \frac{9}{3}=3 \right),{{667}^{\text{th}}}\]term \[\left( \frac{2001}{3}=667 \right)\]are integers. Hence, out of 1000 values of \[\frac{2n+1}{3},\]total number of integral values of\[\frac{2n+1}{3}\] \[=\text{333}+\text{1}=\text{334}\] \[\therefore \]Required probability \[=\frac{334}{1000}=0.334\]
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