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The arithmetic mean of the squares of the first 2n natural numbers is (A) 1/6(2n 1)(4n-1) (B) 1/6(2n-1)(4n-1) (C) 1/6(2n-1)(4n 1) (D) 1/6(2n 1)(4n 1)?
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The arithmetic mean of the squares of the first 2n natural numbers is ...
**Solution:**
To find the arithmetic mean of the squares of the first 2n natural numbers, we first need to find the sum of the squares of these numbers.
Let's denote the sum of the squares as S.
Now, we know that the sum of the squares of the first n natural numbers is given by the formula:
S1 = n(n+1)(2n+1)/6
Using this formula, we can find the sum of the squares of the first 2n natural numbers:
S = (2n)(2n+1)(4n+1)/6
Next, we need to find the arithmetic mean of these squares. The arithmetic mean is equal to the sum divided by the number of terms. In this case, the number of terms is 2n.
So, the arithmetic mean is given by:
(A) 1/6(2n 1)(4n-1)
Therefore, the correct option is (A) 1/6(2n 1)(4n-1).
To find the arithmetic mean of the squares of the first 2n natural numbers, we first need to find the sum of the squares of these numbers.
Let's denote the sum of the squares as S.
Now, we know that the sum of the squares of the first n natural numbers is given by the formula:
S1 = n(n+1)(2n+1)/6
Using this formula, we can find the sum of the squares of the first 2n natural numbers:
S = (2n)(2n+1)(4n+1)/6
Next, we need to find the arithmetic mean of these squares. The arithmetic mean is equal to the sum divided by the number of terms. In this case, the number of terms is 2n.
So, the arithmetic mean is given by:
(A) 1/6(2n 1)(4n-1)
Therefore, the correct option is (A) 1/6(2n 1)(4n-1).
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The arithmetic mean of the squares of the first 2n natural numbers is (A) 1/6(2n 1)(4n-1) (B) 1/6(2n-1)(4n-1) (C) 1/6(2n-1)(4n 1) (D) 1/6(2n 1)(4n 1)?
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