If n numbers are chosen at random from the se | vedclass

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(A) The expression is given by $E = \frac{\sum_{i=1}^n i^2}{\sum_{i=1}^n i}$.Using the standard summation formulas,$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n

Vedclass · Accepted Answer

$0.334$

Vedclass · Answer

(A) The expression is given by $E = \frac{\sum_{i=1}^n i^2}{\sum_{i=1}^n i}$.Using the standard summation formulas,$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^n i = \frac{n(n+1)}{2}$.Substituting these into the expression,we get $E = \frac{n(n+1)(2n+1)/6}{n(n+1)/2} = \frac{2n+1}{3}$.For $E$ to be an integer,$2n+1$ must be divisible by $3$.$2n+1 \equiv 0 \pmod{3} \implies 2n \equiv -1 \equiv 2 \pmod{3} \implies n \equiv 1 \pmod{3}$.In the set $\{1, 2, 3, \dots, 1000\}$,the values of $n$ that satisfy $n \equiv 1 \pmod{3}$ are $1, 4, 7, \dots, 997, 1000$.This is an arithmetic progression with first term $a=1$,last term $l=1000$,and common difference $d=3$.The number of terms $k$ is given by $1000 = 1 + (k-1)3 \implies 999 = 3(k-1) \implies 333 = k-1 \implies k = 334$.The total number of elements in the set is $1000$.The probability is $\frac{334}{1000} = 0.334$.

If $n$ numbers are chosen at random from the set $\{1, 2, 3, \dots, 1000\}$,what is the probability that $\frac{\sum_{i=1}^n i^2}{\sum_{i=1}^n i}$ is an integer?

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