Find the weighted arithmetic mean of the first 'n' natural numbers, the weights being the corresponding numbers.

Concept 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 \(\rm \frac{{(2n + 1)}}{3}\).