Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
across
Group Title
Let's go over the derivation of\[\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}\]
 one year ago
 one year ago
across Group Title
Let's go over the derivation of\[\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}\]
 one year ago
 one year ago

This Question is Closed

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Mathematics group. Uhm.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Does this follow this intuition, somewhat? \[ \sum_{i = 1}^{n} 2i  1= n^2\]?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
\[(k+1)^3=k^3+3k^2+3k+1 \ \ or \ \ (k+1)^3k^3=3k^2+3k+1\] apply sum from \(1\) to \(n\) \[(n+1)^31=\sum_{k=1}^{n} [(k+1)^3k^3]=\sum_{k=1}^{n} [3k^2+3k+1]=3\sum_{k=1}^{n} k^2+3\frac{n(n+1)}{2}+n\]
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
oh this is physics group...o.O.....
 one year ago
See more questions >>>
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.