Same old, same old
Let be a 5th degree monic polynomial such that it satisfies the equations above. Find the value of .
The answer is 279.
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1 solution
Moderator note:
Yes. Method of differences is the correct approach. Can you write the relevant working in your solution?
I used Method of differences for solving this problem we reach up to fourth difference but we need 5th difference.We know
If a polynomial f ( x ) has degree k , then the k th difference is constant.
Furthermore, the k th difference is equal to k ! times the leading coefficient of f ( x ) .
Therefore fifth difference is 5 ! = 1 2 0 .
We complete table using D k + 1 ( n ) = D k ( n + 1 ) − D k ( n )
n 1 2 3 4 5 6 ⋮ f ( n ) 4 9 2 0 4 4 8 8 2 7 9 D 1 ( n ) 5 1 1 2 4 4 4 1 9 1 D 2 ( n ) 6 1 3 2 0 1 4 7 D 3 ( n ) 7 7 1 2 7 D 4 ( n ) 0 1 2 0 D 5 ( n ) … 1 2 0
f ( 6 ) = 2 7 9 .