How to find log of the first n natural numbers? $T(n)= \log 1+ \log 2+ \log 3+\ldots + \log n$ // How to proceed from here?
log ($ 1\times2 \times3\times4\times......\times n$) = log(n!) = log(nn) = O(nlogn)
i think n! can be maped to exponential ....nn ...
so O(nlogn)
you can write log( 1 *2 *3 * ------------------n)
log( n ! ) answer product of 1st n natural number = n !
other one : 1 + 2+ 3+ -------------+ n = n(n+1) /2
same way log 1 + log 2+ -------- log n = $\frac{logn(logn +1 )}{2}$ { but it is not true }
@sumit goyal 1 Anu007 how you you write log1+ log2+ log3+⋯+ logn as log( 1 *2 *3 * ------------------n) as
log( n ! )
hi bhai by log property log a + log b = log ( a.b) so if you have , log a + log b + log c +------- + log n then it is equal to log(a . b . c ------- . n )
and n! = 1.2.3.4............ n @iarnav
@sumit goyal 1 Thanks Bhaijaan :)
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