- #1
eljose
- 492
- 0
Let,s suppose we want to do this sum:
[tex] 1+2^{m}+3^{m}+...+n^{m} [/tex] n finite
then we could use the property of the differences:
[tex] \sum_{n=0}^{n}(y(k)-y(k-1))=y(n)-y(0) [/tex]
so for any function of the form f(x)=x^{m} m integer you need to solve:
[tex] y(n)-y(n-1)=n^{m} [/tex] 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?..
[tex] 1+2^{m}+3^{m}+...+n^{m} [/tex] n finite
then we could use the property of the differences:
[tex] \sum_{n=0}^{n}(y(k)-y(k-1))=y(n)-y(0) [/tex]
so for any function of the form f(x)=x^{m} m integer you need to solve:
[tex] y(n)-y(n-1)=n^{m} [/tex] 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?..