A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
The sum of \(n\) real numbers is zero and the sum of their pairwise products is also zero. Prove that the sum of the cubes of the numbers is zero.
 2 years ago
The sum of \(n\) real numbers is zero and the sum of their pairwise products is also zero. Prove that the sum of the cubes of the numbers is zero.

This Question is Closed

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2\[(a _{1}+a _{2}+a _{3}+...+a _{n})^2=a _{1}^2+a _{2}^2+a _{3}^2+...+a _{n}^2+2(a _{1}a _{2}+a _{1}a _{3}+...+a _{n1}a _{n})\\ a _{1}^2+a _{2}^2+a _{3}^2+...+a _{n}^2=(a _{1}+a _{2}+a _{3}+...+a _{n})2(a _{1}a _{2}+a _{1}a _{3}+...+a _{n1}a _{n})=0 \]

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2then \[a_{1}=a_{2}=a_{3}=...=a_{n}=0 \\ so \\ \] sum of the cubes of the numbers is zero

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0I see. This wasn't that hard. Only my brain stopped thinking.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.