Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course


 

Problem

If k is an odd positive integer, prove that for any integer \mathbf{ n ge 1 , 1^k + 2^k + \cdots + n^k } is divisible by \mathbf{ \frac {n(n+1)}{2} }

Discussion

We write the given expression in two ways:

\mathbf{ S= 1^k + 2^k + \cdots + n^k }
\mathbf{ S =n^k + (n-1)^k + \cdots + 1^k }

This implies

\mathbf{ 2S = (1^k + n^k ) + (2^k + (n-1)^k ) + \cdots + (n^k + 1^k) }
Since k is odd, we know \mathbf{ a^k + b^k = (a+b)(a^{k-1} - a^{k-2} b + \cdots + b^{k-1}) } , that is \mathbf{ a+b} divides \mathbf{a^k + b^k }

Applying this to \mathbf{ 2S = (1^k + n^k ) + (2^k + (n-1)^k ) + \cdots + (n^k + 1^k) } we have (n+1) divides \mathbf{ 1^k + n^k } , \mathbf{ (2 + n-1) =(n+1) } divides \mathbf{(2^k + (n-1)^k)} and so on. Hence we can take n+1 common from each bracket, leading us to the following expression:

\mathbf{ 2S = (n+1)\times(\text{some constant}) \implies S = \frac{(n+1)}{2} \times(\text{some constant}) } . This implies \mathbf{ \frac{(n+1)}{2} } divides S if n+1 is even, other wise n+1 divides S.

Now we show n (or n/2) divides S (when is odd or even respectively). To show this we write

S= \mathbf{ 1^k + 2^k + \cdots + n^k }
\mathbf{ S =n^k + (n-1)^k + \cdots + 1^k }

Again \mathbf{ 2S =n^k + (1^k+ (n-1)^k) + (2^k + (n-2)^k) \cdots + ((n-1)^k + 1^k) + n^k }
Since k is odd n divides \mathbf{ a^k + (n-a)^k } for all a from 1 to n. Hence we can take n as common and have:

\mathbf{ 2S = (n)\times(\text{some constant}) \implies S = \frac{(n)}{2} \times(\text{some constant}) } . This implies \mathbf{ \frac{(n)}{2} } divides S if n is even, other wise n divides S.

Now gcd of (n, n+1) = 1. Hence \mathbf{ \frac{n(n+1)}{2} } divides S.

Proved