Problem Statement:
Let F:R→R be a non-constant function that is three times differentiable. If F(1+1/n)=0 for all integers n, then what is the value of F'(1)? Provide a detailed explanation.

Solution:

Given:
- F:R→R is a non-constant function
- F is three times differentiable
- F(1+1/n)=0 for all integers n

Approach:
To find the value of F'(1), we can use the concept of Taylor's expansion and the fact that F(1+1/n)=0 for all integers n.

Step 1: Taylor's Expansion:
Taylor's expansion allows us to approximate a function using a polynomial. We can expand F(x) around x=1 using Taylor's expansion up to the third derivative:

F(x) = F(1) + F'(1)(x-1) + (F''(1)/2)(x-1)^2 + (F'''(c)/6)(x-1)^3

where c is some value between 1 and x.

Step 2: F(1+1/n)=0:
Given that F(1+1/n)=0 for all integers n, we can substitute x=1+1/n into the Taylor's expansion:

0 = F(1) + F'(1)(1+1/n-1) + (F''(1)/2)(1+1/n-1)^2 + (F'''(c)/6)(1+1/n-1)^3

Simplifying the equation, we get:

0 = F(1) + F'(1)/n + (F''(1)/2)(1/n^2) + (F'''(c)/6)(1/n^3)

Step 3: Taking the Limit:
Now, we take the limit as n approaches infinity on both sides of the equation:

lim(n→∞) 0 = lim(n→∞) [F(1) + F'(1)/n + (F''(1)/2)(1/n^2) + (F'''(c)/6)(1/n^3)]

As n approaches infinity, the terms with 1/n^2 and 1/n^3 tend to 0. Therefore, we are left with:

0 = F(1) + 0 + 0

Simplifying further, we find:

F(1) = 0

Step 4: F'(1):
Substituting F(1) = 0 into the Taylor's expansion, we have:

F(x) = 0 + F'(1)(x-1) + (F''(1)/2)(x-1)^2 + (F'''(c)/6)(x-1)^3

Now, we differentiate both sides of the equation with respect to x:

F'(x) = F'(1) + (F''(1))(x-1) + (F'''(c)/2)(x-1)^2

Substituting x=1, we get:

F'(1) = F'(1) + (F''(1))(1-1) + (F'''(c)/