Question 112

What is the average of the cubes of the first 13 natural numbers?

Solution

Sum of the cubes of the first $$n$$ natural numbers = $$[\frac{n(n+1)}{2}]^2$$

=> Sum of the cubes of the first 13 natural numbers = $$[\frac{13(13+1)}{2}]^2$$

=> $$(13)^2\times49$$

$$\therefore$$ Required average = $$\frac{(13)^2\times49}{13}$$

= $$13\times49=637$$

=> Ans - (D)

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