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