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