Question

Solutions

For positive integers n; define
f (n)=1^n + 2^n-1 + 3^n-2 +.......... (n-2)^3 + (n-1)^2 + n:
What is the minimum of f (n + 1)=f (n)?

Answer 1

For n = 1, 2,3, 4, 5, 6, f (n + 1) / f (n) = 3; 8/3; 22/8; 65/22; 209/65; 732/209; respectively. The minimum of these is 8/3: For n > 6; we will show f (n + 1)=f (n) > 3 > 8/3:

This follows from

f (n + 1)

>1^n+1 + 2^n + 3^n-1 + 4^n-2 + 5^n-3 + 6^n-4 +.........+ (n -1)^3 + n^2

>1^n+1 + 2^n + 3^n-1 + 4^n-2 + 5^n-3 + 3(6^n-5 + ...... + (n -1)^2 + n)

=1^n+1 + ....... + 5^n-3 + 3(f (n)- 1^n - 2^n-1 + 3^n-2 + 4^n-3 + 5^n-4 )

=3f (n) + 2(5^n-4 - 1) + 2^n-1 (2^n-5 -1) > 3f (n):

Therefore, 8/3 is the answer.