Question
Download Solution PDFFind the weighted arithmetic mean of the first 'n' natural numbers, the weights being the corresponding numbers.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept used:
1. Sum of the squares of first n natural numbers = \(\frac {n(n +1)(2n +1)}{6}\)
2. Sum of n natural numbers = \(\frac {n(n +1)}{2}\)
Calculation:
First n natural numbers = 1, 2, 3, ...., n
Their corresponding weights respectively = 1, 2, 3, ...., n
Now, the weighted arithmetic mean of the first 'n' natural numbers
⇒ \(\frac {1 \times 1 + 2 \times 2 +3 \times 3 + ..... + n \times n}{1 + 2 + 3 + .... + n}\)
⇒ \(\frac {1^2 + 2^2 + 3^2 + .... + n^2}{1 + 2 + 3 + .... + n}\)
⇒ \(\frac {\frac {n(n +1)(2n +1)}{6}} {\frac {n(n +1)}{2}}\)
⇒ \(\rm \frac{{(2n + 1)}}{3}\)
∴ The weighted arithmetic mean of the first 'n' natural numbers is \(\rm \frac{{(2n + 1)}}{3}\).
Last updated on Jul 10, 2025
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