IIT JEE , AIEEE Forum - Mathematics , Algebra

anchitsaini

i had learnt of this proof somewhere(most probably over the internet)--

lets start with proof for sum of divisors of p^k where p is a prime number

sum of its divisors should be (p^k+1)/ (p-1)

proof--
divisors of p^k = 1,p,p², ......p^k
hence sum of divisors is a GP =(p^k+1-1) / (p-1)

now if a number can be represented as

n = p₁^r1 . ...... p_k^rk

hence sum of its divisors would be --

( p₁^r1+1 - 1)/ (p₁-1) ... ......( p_k^rk+1 - 1) / (p_k-1)

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