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frx

  • 2 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...

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  1. calculusfunctions
    • 2 years ago
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    Have you ever learned mathematical induction?

  2. AbhimanyuPudi
    • 2 years ago
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    Mathematical induction will do..

  3. frx
    • 2 years ago
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    Yes, but i don't know how to use it in this case

  4. AbhimanyuPudi
    • 2 years ago
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    Its the same..nothing different..try it on paper

  5. frx
    • 2 years ago
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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?

  6. calculusfunctions
    • 2 years ago
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    Do you also understand that\[\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{ n! }{ r!(n -r)! }\]

  7. frx
    • 2 years ago
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    Yes I do

  8. AbhimanyuPudi
    • 2 years ago
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    Nothing different..choose k..

  9. calculusfunctions
    • 2 years ago
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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

  10. frx
    • 2 years ago
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    Ok, thanks then there should be no problem i think :)

  11. AbhimanyuPudi
    • 2 years ago
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    Yes..: )

  12. Zarkon
    • 2 years ago
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    A little \(\LaTeX\) tip you can use {n\choose r} for \[{n\choose r}\]

  13. Zarkon
    • 2 years ago
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    \[\binom{n}{r}\] or \binom{n}{r}

  14. frx
    • 2 years ago
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    Great, didn't find that in the toolbar but will remember it further on :)

  15. frx
    • 2 years ago
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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?

  16. helder_edwin
    • 2 years ago
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    assuming that (for k>1) \[ \large \sum_{i=2}^{k+1}\binom{i}{2}>\frac{k^3}{6} \] then \[ \large \sum_{i=2}^{k+2}\binom{i}{2}>\frac{(k+1)^3}{6} \]

  17. helder_edwin
    • 2 years ago
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    so \[ \large \sum_{i= 2}^{k+2}=\sum_{i=2}^{k+1}\binom{i}{2}+\binom{k+2}{2} >\frac{k^3}{6}+\frac{(k+2)!}{2!k!} \]

  18. helder_edwin
    • 2 years ago
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    \[ \large =\frac{k!k^3+3(k+2)!}{6k!} \]

  19. helder_edwin
    • 2 years ago
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    \[ \large =\frac{k^3+3(k+2)(k+1)(k-1)!}{6} \]

  20. frx
    • 2 years ago
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    How did you do that last simplification?

  21. helder_edwin
    • 2 years ago
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    \[ \large =\frac{k^3+3(k^2+3k+2)(k-1)!}{6} \]

  22. frx
    • 2 years ago
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    Dividing 3(k+2)! with k!

  23. helder_edwin
    • 2 years ago
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    i factored k! from the numerator

  24. helder_edwin
    • 2 years ago
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    remember this \[ \large n!=n(n-1)! \]

  25. helder_edwin
    • 2 years ago
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    i made a mistake

  26. helder_edwin
    • 2 years ago
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    it should be \[ \large =\frac{k![k^3+3(k+2)(k+1)]}{6k!}=\frac{k^3+3(k+2)(k+1)}{6} \]

  27. frx
    • 2 years ago
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    Oh now I see it!

  28. frx
    • 2 years ago
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    Thank you so much, really appreciate it :D

  29. helder_edwin
    • 2 years ago
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    \[ \large =\frac{k^3+3k^2+9k+6}{6} \]

  30. frx
    • 2 years ago
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    Obviously i didn't still don't get the part where you factor the k! out from 3(k+2)!

  31. helder_edwin
    • 2 years ago
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    \[ \large \frac{(k^3+3k^2+3k+1)+6k+5}{6} \]

  32. helder_edwin
    • 2 years ago
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    using the (recursive) definition of factorial \[ \large (k+2)!=(k+2)(k+1)!=(k+2)(k+1)k! \]

  33. helder_edwin
    • 2 years ago
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    now resuming: since k>1 \[ \large =\frac{(k+1)^3+6k+5}{6}>\frac{(k+1)^3}{6} \]

  34. helder_edwin
    • 2 years ago
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    q.e.d.

  35. frx
    • 2 years ago
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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?

  36. frx
    • 2 years ago
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    110% got it! Wonderful! Once again, thank you!

  37. helder_edwin
    • 2 years ago
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    u r welcome

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