A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Show that:
\[\left(\begin{matrix}2 \\ 2\end{matrix}\right)+\left(\begin{matrix}3 \\ 2\end{matrix}\right)+\left(\begin{matrix}4 \\ 2\end{matrix}\right)+ ....+\left(\begin{matrix}n+1 \\ 2\end{matrix}\right) > \frac{ n ^{3} }{ 6}\]
is true for all n=2,3,4...
anonymous
 4 years ago
Show that: \[\left(\begin{matrix}2 \\ 2\end{matrix}\right)+\left(\begin{matrix}3 \\ 2\end{matrix}\right)+\left(\begin{matrix}4 \\ 2\end{matrix}\right)+ ....+\left(\begin{matrix}n+1 \\ 2\end{matrix}\right) > \frac{ n ^{3} }{ 6}\] is true for all n=2,3,4...

This Question is Closed

calculusfunctions
 4 years ago
Best ResponseYou've already chosen the best response.1Have you ever learned mathematical induction?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Mathematical induction will do..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, but i don't know how to use it in this case

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Its the same..nothing different..try it on paper

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Step 1: try out for n=2, Step 2: try out for n=k Step 3 try for n=k+1, so it's nothing different even though it is n+1 choose k?

calculusfunctions
 4 years ago
Best ResponseYou've already chosen the best response.1Do you also understand that\[\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{ n! }{ r!(n r)! }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nothing different..choose k..

calculusfunctions
 4 years ago
Best ResponseYou've already chosen the best response.1Step 1: Try for n = 1 Step 2: Assume true for n = k Step 3: prove true for n = k + 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, thanks then there should be no problem i think :)

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0A little \(\LaTeX\) tip you can use {n\choose r} for \[{n\choose r}\]

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0\[\binom{n}{r}\] or \binom{n}{r}

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Great, didn't find that in the toolbar but will remember it further on :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ {2\choose 2} + {3\choose 2}+ {4\choose 2}+...++ {k+2\choose 2} > \frac{ (k+1)^{3} }{ 6}\] \[\frac{ k ^{3} }{ 6 }+{k+2\choose 2} = \frac{ k ^{3} }{ 6 } + \frac{ (k+2)! }{ 2!k!}=\frac{ 2!k!k ^{3}+6(k+2)! }{ 2!k!6 }=\frac{ k!k ^{3}+3(k+2)! }{ k!6 }\] I don't know how to simplify the resy, could someone please help me?

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1assuming that (for k>1) \[ \large \sum_{i=2}^{k+1}\binom{i}{2}>\frac{k^3}{6} \] then \[ \large \sum_{i=2}^{k+2}\binom{i}{2}>\frac{(k+1)^3}{6} \]

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1so \[ \large \sum_{i= 2}^{k+2}=\sum_{i=2}^{k+1}\binom{i}{2}+\binom{k+2}{2} >\frac{k^3}{6}+\frac{(k+2)!}{2!k!} \]

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1\[ \large =\frac{k!k^3+3(k+2)!}{6k!} \]

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1\[ \large =\frac{k^3+3(k+2)(k+1)(k1)!}{6} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How did you do that last simplification?

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1\[ \large =\frac{k^3+3(k^2+3k+2)(k1)!}{6} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Dividing 3(k+2)! with k!

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1i factored k! from the numerator

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1remember this \[ \large n!=n(n1)! \]

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1it should be \[ \large =\frac{k![k^3+3(k+2)(k+1)]}{6k!}=\frac{k^3+3(k+2)(k+1)}{6} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you so much, really appreciate it :D

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1\[ \large =\frac{k^3+3k^2+9k+6}{6} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Obviously i didn't still don't get the part where you factor the k! out from 3(k+2)!

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1\[ \large \frac{(k^3+3k^2+3k+1)+6k+5}{6} \]

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1using the (recursive) definition of factorial \[ \large (k+2)!=(k+2)(k+1)!=(k+2)(k+1)k! \]

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1now resuming: since k>1 \[ \large =\frac{(k+1)^3+6k+5}{6}>\frac{(k+1)^3}{6} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh so you're just saying that just as 2! = 2*1 the number before k+2 must have been k+1, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0110% got it! Wonderful! Once again, thank you!
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.