Prove that $\sum_{i \in A} i k_i = \sum_{i \in B} i \Rightarrow \sum_{i \in A} i^2 k_i \neq \sum_{i \in B} i^2$

Asked Oct 14 '20 at 13:24

Active Oct 14 '20 at 13:26

Viewed 39 times

I would like to prove that, given two disjoint sets of integers A and B (with elements strictly greater than 1), and a sequence of integers $k_i > 0$, then $\sum_{i \in A} i k_i = \sum_{i \in B} i \Rightarrow \sum_{i \in A} i^2 k_i \neq \sum_{i \in B} i^2$

edited Oct 14 '20 at 13:26

asked Oct 14 '20 at 13:24

weghiuop

2

Not true for $A=B$ and $k_i \equiv 1$. Please show some context and work. – player3236 Oct 14 '20 at 13:26
You are right, I added some constraints – weghiuop Oct 14 '20 at 13:28

Prove that $\sum_{i \in A} i k_i = \sum_{i \in B} i \Rightarrow \sum_{i \in A} i^2 k_i \neq \sum_{i \in B} i^2$

0 Answers0