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\[\left(\begin{matrix}2 \\ 2\end{matrix}\right)+\left(\begin{matrix}3 \\ 2\end{matrix}\right)+\left(\begin{matrix}4 \\ 2\end{matrix}\right)+ ....+\left(\begin{matrix}n+1 \\ 2\end{matrix}\right) > \frac{ n ^{3} }{ 6}\]
is true for all n=2,3,4...
 one year ago
 one year 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...
 one year ago
 one year ago

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calculusfunctionsBest ResponseYou've already chosen the best response.1
Have you ever learned mathematical induction?
 one year ago

AbhimanyuPudiBest ResponseYou've already chosen the best response.1
Mathematical induction will do..
 one year ago

frxBest ResponseYou've already chosen the best response.0
Yes, but i don't know how to use it in this case
 one year ago

AbhimanyuPudiBest ResponseYou've already chosen the best response.1
Its the same..nothing different..try it on paper
 one year ago

frxBest ResponseYou've already chosen the best response.0
Step 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?
 one year ago

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

AbhimanyuPudiBest ResponseYou've already chosen the best response.1
Nothing different..choose k..
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.1
Step 1: Try for n = 1 Step 2: Assume true for n = k Step 3: prove true for n = k + 1
 one year ago

frxBest ResponseYou've already chosen the best response.0
Ok, thanks then there should be no problem i think :)
 one year ago

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

ZarkonBest ResponseYou've already chosen the best response.0
\[\binom{n}{r}\] or \binom{n}{r}
 one year ago

frxBest ResponseYou've already chosen the best response.0
Great, didn't find that in the toolbar but will remember it further on :)
 one year ago

frxBest 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?
 one year ago

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

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

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

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

frxBest ResponseYou've already chosen the best response.0
How did you do that last simplification?
 one year ago

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

helder_edwinBest ResponseYou've already chosen the best response.1
i factored k! from the numerator
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.1
remember this \[ \large n!=n(n1)! \]
 one year ago

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

frxBest ResponseYou've already chosen the best response.0
Thank you so much, really appreciate it :D
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.1
\[ \large =\frac{k^3+3k^2+9k+6}{6} \]
 one year ago

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

helder_edwinBest ResponseYou've already chosen the best response.1
\[ \large \frac{(k^3+3k^2+3k+1)+6k+5}{6} \]
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.1
using the (recursive) definition of factorial \[ \large (k+2)!=(k+2)(k+1)!=(k+2)(k+1)k! \]
 one year ago

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

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

frxBest ResponseYou've already chosen the best response.0
110% got it! Wonderful! Once again, thank you!
 one year ago
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