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The number N=6^(log(10)40)*5^(log(10)36)...

The number `N=6^(log_(10)40)*5^(log_(10)36)` is a natural number. Then sum of digits of N is :

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The value of 6^(log_10 40)*5^(log_10 36) is

Prove that log_n(n+1)>log_(n+1)(n+2) for any natural number n > 1.

Knowledge Check

  • The sum of n odd natural numbers is

    A
    `n^(2)`
    B
    `2n`
    C
    `(n+1)/(2)`
    D
    `n^(2)+1`

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