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amistre64Best ResponseYou've already chosen the best response.2
sounds inductiony to me
 one year ago

cinarBest ResponseYou've already chosen the best response.0
actually I found general formula for sum, but I am not sure if I can use it as a proof..
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
can you work it up to the formula? then once a general formula is "determined" you might be able to prove it by induction
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
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?
 one year ago

cinarBest ResponseYou've already chosen the best response.0
here it is.. http://imageshack.us/photo/myimages/513/17165198.png/
 one year ago

cinarBest ResponseYou've already chosen the best response.0
You see the poly is always m+1 degree when expression is m degree
 one year ago

cinarBest ResponseYou've already chosen the best response.0
this the website http://www.trans4mind.com/personal_development/mathematics/series/sumsBernoulliNumbers.htm
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
that looks like a binomial expansion to me
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
\[\sum(n+1)^k=\binom{k}{0}n^k1^{0}+\binom{k}{1}n^{k1}1^1+\binom{k}{2}n^{k2}1^2+...+\binom{k}{k}n^{kk}1^k\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
n=0,1,2,3,... produces the same summation i bleieve as:1^k + 2^k + 3^k + ...
 one year ago

cinarBest ResponseYou've already chosen the best response.0
can we use this notation\[\sum_{n=0 }^k(n+1)^k=\binom{k}{n}n^k1^{0}+\binom{k}{n+1}n^{k1}1^1+\binom{k}{n+2}n^{k2}1^2+...+\binom{k}{n+k1}n^{kk}1^k\]
 one year ago

cinarBest ResponseYou've already chosen the best response.0
I guess n should be start from 1
 one year ago

cinarBest ResponseYou've already chosen the best response.0
can we use this notation\[\sum_{i=1 }^n i^k=1^k+2^k+......+n^k=?\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
something that bugs me about what i posted is that if n=0, we run into a 0^0 case ....
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
im not read up n the bernuolli numbers enough to comment if thats a suitable "showing" or not
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
i^1; show a second degree poly 1,3,6,10,15 2 3 4 5 1 1 \[1+2n+\frac{n(n1)}{2!}~:~n=0,1,2,3,...\] 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 \[1+4n+5\frac{n(n1)}{2!}+2\frac{n(n1)(n2)}{3!}~:~n=0,1,2,3,...\]is a 3rd degree poly
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
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
 one year ago

cinarBest ResponseYou've already chosen the best response.0
thanks, you helped me a lot..
 one year ago
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