I conjectured that $ \left(\sum _ {i_k = 1} ^ n\right) ^ k i = {_{n+k} C _ {k+1}} $ where $n , k \in \mathbb N$ and $ i_k = \left(\sum _ {i_{(k-1)} = 1} ^ n\right) ^ {(k-1)} i $. The notation I adopted is similar with the notation used for composition of function $ f^2 (x) = f(f(x)) \neq (f(x))^2$. Now if for a specific natural number $ k$ is given, then the conjecture can be proved using mathematical induction. But how can we prove for any $k$. Thank you.
A composition on finite sequence
0
$\begingroup$
sequences-and-series
-
1I do not follow your notation. Your explanation "The notation I adopted is similar with the notation used for composition of function $f^2(x)=f(f(x))$" is also a little cryptic. Can you clarify what you mean? [Perhaps explicitly writing out what you mean for $k=2$ might help.] – 2011-11-19
-
0Suppose k = 2. $(\sum _ {i= 1} ^ n) ^ k i = _{(n+k)} C _ {(k+1)} $ will become $ \sum_{i= 1} ^ n \sum_{i= 1} ^ n i =\frac {(n+2)!}{3! (n-1)!} $. I hope it helps. – 2011-11-19
-
1But the i for the second summation represent the value of the first summation – 2011-11-19
-
1@Srivatsan: my guess (ken please confirm or deny) is $\sum_{i= 1} ^ n \sum_{i= 1} ^ n i =\sum_{i= 1} ^ n \frac{n(n+1)}{2}$ (which should be) $\sum_{i= 1} ^ n \frac{i(i+1)}{2}=\frac{n(n+1)(2n+1}{12}+\frac{n(n+1)}{2}$ but that doesn't equal $\frac {(n+2)!}{3! (n-1)!}$ ken, you should use different dummy variables in the nested sums-one should use $j$ for example. – 2011-11-19
-
0Yes. thank you.@ Ross Millikan: Thank you for the notation and the answer. :) – 2011-11-19