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cinar
discrete math question..need some help..
sounds inductiony to me
actually I found general formula for sum, but I am not sure if I can use it as a proof..
can you work it up to the formula? then once a general formula is "determined" you might be able to prove it by induction
of course, you might have a better notion of what you need. What are your thoughts on it? what have you been thinking would work?
here it is.. http://imageshack.us/photo/my-images/513/17165198.png/
You see the poly is always m+1 degree when expression is m degree
this the website http://www.trans4mind.com/personal_development/mathematics/series/sumsBernoulliNumbers.htm
that looks like a binomial expansion to me
\[\sum(n+1)^k=\binom{k}{0}n^k1^{0}+\binom{k}{1}n^{k-1}1^1+\binom{k}{2}n^{k-2}1^2+...+\binom{k}{k}n^{k-k}1^k\]
n=0,1,2,3,... produces the same summation i bleieve as:1^k + 2^k + 3^k + ...
can we use this notation\[\sum_{n=0 }^k(n+1)^k=\binom{k}{n}n^k1^{0}+\binom{k}{n+1}n^{k-1}1^1+\binom{k}{n+2}n^{k-2}1^2+...+\binom{k}{n+k-1}n^{k-k}1^k\]
I guess n should be start from 1
can we use this notation\[\sum_{i=1 }^n i^k=1^k+2^k+......+n^k=?\]
something that bugs me about what i posted is that if n=0, we run into a 0^0 case ....
im not read up n the bernuolli numbers enough to comment if thats a suitable "showing" or not
i^1; show a second degree poly 1,3,6,10,15 2 3 4 5 1 1 \[1+2n+\frac{n(n-1)}{2!}~:~n=0,1,2,3,...\] is a 2nd degree poly i^2; show a third degree poly 1 5 14 30 55 4 9 16 25 5 7 9 2 2 \[1+4n+5\frac{n(n-1)}{2!}+2\frac{n(n-1)(n-2)}{3!}~:~n=0,1,2,3,...\]is a 3rd degree poly
if we can construct this polynomial as a pattern; then we can try to induct that pattern to the k+1th term or would we have to prove that each substructure is inductable? this is the part i hate about proofs, you never know when you have to reinvent the wheel
thanks, you helped me a lot..