anonymous
  • anonymous
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...
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
calculusfunctions
  • calculusfunctions
Have you ever learned mathematical induction?
anonymous
  • anonymous
Mathematical induction will do..
anonymous
  • anonymous
Yes, but i don't know how to use it in this case

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anonymous
  • anonymous
Its the same..nothing different..try it on paper
anonymous
  • anonymous
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?
calculusfunctions
  • calculusfunctions
Do you also understand that\[\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{ n! }{ r!(n -r)! }\]
anonymous
  • anonymous
Yes I do
anonymous
  • anonymous
Nothing different..choose k..
calculusfunctions
  • calculusfunctions
Step 1: Try for n = 1 Step 2: Assume true for n = k Step 3: prove true for n = k + 1
anonymous
  • anonymous
Ok, thanks then there should be no problem i think :)
anonymous
  • anonymous
Yes..: )
Zarkon
  • Zarkon
A little \(\LaTeX\) tip you can use {n\choose r} for \[{n\choose r}\]
Zarkon
  • Zarkon
\[\binom{n}{r}\] or \binom{n}{r}
anonymous
  • anonymous
Great, didn't find that in the toolbar but will remember it further on :)
anonymous
  • anonymous
\[ {2\choose 2} + {3\choose 2}+ {4\choose 2}+...++ {k+2\choose 2} > \frac{ (k+1)^{3} }{ 6}\] \[\frac{ k ^{3} }{ 6 }+{k+2\choose 2} = \frac{ k ^{3} }{ 6 } + \frac{ (k+2)! }{ 2!k!}=\frac{ 2!k!k ^{3}+6(k+2)! }{ 2!k!6 }=\frac{ k!k ^{3}+3(k+2)! }{ k!6 }\] I don't know how to simplify the resy, could someone please help me?
helder_edwin
  • helder_edwin
assuming that (for k>1) \[ \large \sum_{i=2}^{k+1}\binom{i}{2}>\frac{k^3}{6} \] then \[ \large \sum_{i=2}^{k+2}\binom{i}{2}>\frac{(k+1)^3}{6} \]
helder_edwin
  • helder_edwin
so \[ \large \sum_{i= 2}^{k+2}=\sum_{i=2}^{k+1}\binom{i}{2}+\binom{k+2}{2} >\frac{k^3}{6}+\frac{(k+2)!}{2!k!} \]
helder_edwin
  • helder_edwin
\[ \large =\frac{k!k^3+3(k+2)!}{6k!} \]
helder_edwin
  • helder_edwin
\[ \large =\frac{k^3+3(k+2)(k+1)(k-1)!}{6} \]
anonymous
  • anonymous
How did you do that last simplification?
helder_edwin
  • helder_edwin
\[ \large =\frac{k^3+3(k^2+3k+2)(k-1)!}{6} \]
anonymous
  • anonymous
Dividing 3(k+2)! with k!
helder_edwin
  • helder_edwin
i factored k! from the numerator
helder_edwin
  • helder_edwin
remember this \[ \large n!=n(n-1)! \]
helder_edwin
  • helder_edwin
i made a mistake
helder_edwin
  • helder_edwin
it should be \[ \large =\frac{k![k^3+3(k+2)(k+1)]}{6k!}=\frac{k^3+3(k+2)(k+1)}{6} \]
anonymous
  • anonymous
Oh now I see it!
anonymous
  • anonymous
Thank you so much, really appreciate it :D
helder_edwin
  • helder_edwin
\[ \large =\frac{k^3+3k^2+9k+6}{6} \]
anonymous
  • anonymous
Obviously i didn't still don't get the part where you factor the k! out from 3(k+2)!
helder_edwin
  • helder_edwin
\[ \large \frac{(k^3+3k^2+3k+1)+6k+5}{6} \]
helder_edwin
  • helder_edwin
using the (recursive) definition of factorial \[ \large (k+2)!=(k+2)(k+1)!=(k+2)(k+1)k! \]
helder_edwin
  • helder_edwin
now resuming: since k>1 \[ \large =\frac{(k+1)^3+6k+5}{6}>\frac{(k+1)^3}{6} \]
helder_edwin
  • helder_edwin
q.e.d.
anonymous
  • anonymous
Oh so you're just saying that just as 2! = 2*1 the number before k+2 must have been k+1, right?
anonymous
  • anonymous
110% got it! Wonderful! Once again, thank you!
helder_edwin
  • helder_edwin
u r welcome

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