A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

giovkast
 2 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{k=1}^{n} k^2 = 1/6n(n + 1)(2n+1)\]

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.0first step: n = 1: \[\sum_{k=1}^{1}k^2=1^2 = 1,\quad (1/6)(1)(2)(3) = 1\]

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.0step 2: induction process. assume true for n=s \[\sum_{k=1}^{s}=1^2+2^2+\cdots +s^2=(1/6) s(s+1)(2s+1)\] show the formula is true for n = s+1 \[\sum_{k=1}^{s+1}=1^2+2^2+\cdots + s^2 + (s+1)^2\]\[=\left[ (1/6)s(s+1)(2s+1) \right]+(s+1)^2\]

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.0simplify the expressions, then factor. if you get \[(1/6)(s+1)(s+2)(2s+3)\] then you're in the right track.

giovkast
 2 years ago
Best ResponseYou've already chosen the best response.0\[1/3n^3+3/2n2+3/2n+1\]

giovkast
 2 years ago
Best ResponseYou've already chosen the best response.0i got this, don't know where i went wrong

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.2you need to add \[\frac{s(s+1)(2s+1)}{6}+\frac{(s+1)^2}{1}=\frac{s(s+1)(2s+1)+6(s+1)^2}{6}\] \[=\frac{(s+1)[s(2s+1)+6(s+1)]}{6}=\frac{(s+1)(2s^2+7s+6)}{6}=\frac{(s+1)(2s+3)(s+2)}{6}\] we need to show that this is the same as n=s+1 in the original ie \[\frac{(\color{red}{s+1})(2\color{red}{(s+1)}+1)(\color{red}{s+1}+1)}{6}\]TRUE
Ask your own question
Sign UpFind 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.