eljose
- 484
- 0
Let,s suppose we want to do this sum:
1+2^{m}+3^{m}+...+n^{m} n finite
then we could use the property of the differences:
\sum_{n=0}^{n}(y(k)-y(k-1))=y(n)-y(0)
so for any function of the form f(x)=x^{m} m integer you need to solve:
y(n)-y(n-1)=n^{m} i don,t know how to solve
it..
i have tried the ansatz y(n)=K(n) with K(n) a Polynomial of degree m+1 but i don,t get the usual results for the sum..could someone help?..
1+2^{m}+3^{m}+...+n^{m} n finite
then we could use the property of the differences:
\sum_{n=0}^{n}(y(k)-y(k-1))=y(n)-y(0)
so for any function of the form f(x)=x^{m} m integer you need to solve:
y(n)-y(n-1)=n^{m} i don,t know how to solve
it..

