Question

The sum of the cubes of the first n natural numbers is 2025, then find the value of n

Answer 1

1² + 2² + 3² + … + n² = 285

`("n"("n" + 1)(2"n" + 1))/6` = 285   ...(1)

1³ + 2³ + 3³ + … + n³ = 2025

`[("n"("n" + 1))/2]^2` = 2025

`("n"("n" + 1))/2 = sqrt(2025)`

⇒ `("n"("n" + 1))/2` = 45

Substitute the value of `("n"("n" + 1))/2` in (1)

`(45(2"n" + 1))/3` = 285

(2n + 1) = `(285 xx 3)/45 = 285/15` = 19

2n + 1 = 19 ⇒ 2n = 19 – 1

2n = 18 ⇒ n = `18/2` = 9

The value of n is 9.