find the mean of squares of first n natural numbers
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1
Step-by-step explanation:
The average of squares of first n natural numbers = (n+1)(2n+1)/6. The average of cubes of first n natural numbers = n(n+1)2/4. ... The average of even numbers from 1 to n = (last even number+2)/2. The average of squares of first n odd numbers = (2n+1)(2n-1)/3.
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Answered by
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The first n natural numbers are 1,2,3,...,n
Sum of squares of n natural numbers is 1
2
+2
2
+3
2
+...+n
2
We know that sum of squares of first n natural numbers is given by
6
n(n+1)(2n+1)
Mean is the average of the values of the data set.
Therefore, mean of the squares of first n natural numbers is
n
1
2
+2
2
+3
2
+...+n
2
=
6n
n(n+1)(2n+1)
=
6
(n+1)(2n+1)
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