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If the product of `n` positive numbers is `n^n`. Then their sum is

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Knowledge Check

  • The product of n positive numbers is 1, then their sum is a positive integer, that is

    A
    equal to 1
    B
    equal to `n + n^(2)`
    C
    divisible by `n`
    D
    never less than `n`

  • If the sum of first n positive integers is 1/5 times the sum of their square then n equals

    A
    5
    B
    6
    C
    7
    D
    8

  • The product of first n odd natural numbers equals

    A
    `""^(2n)C_(n)xx""^(n)P_(n)`
    B
    `(1/2)^(n)xx""^(2n)C_(n)xx""^(n)P_(n)`
    C
    `(1/4)xx""^(2n)C_(n)xx""^(2n)P_(n)`
    D
    `""^(n)C_(n)xx""^(n)P_(n)`

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    The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.

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