## anonymous 3 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...

1. calculusfunctions

Have you ever learned mathematical induction?

2. anonymous

Mathematical induction will do..

3. anonymous

Yes, but i don't know how to use it in this case

4. anonymous

Its the same..nothing different..try it on paper

5. 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?

6. calculusfunctions

Do you also understand that$\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{ n! }{ r!(n -r)! }$

7. anonymous

Yes I do

8. anonymous

Nothing different..choose k..

9. calculusfunctions

Step 1: Try for n = 1 Step 2: Assume true for n = k Step 3: prove true for n = k + 1

10. anonymous

Ok, thanks then there should be no problem i think :)

11. anonymous

Yes..: )

12. Zarkon

A little $$\LaTeX$$ tip you can use {n\choose r} for ${n\choose r}$

13. Zarkon

$\binom{n}{r}$ or \binom{n}{r}

14. anonymous

Great, didn't find that in the toolbar but will remember it further on :)

15. 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?

16. helder_edwin

assuming that (for k>1) $\large \sum_{i=2}^{k+1}\binom{i}{2}>\frac{k^3}{6}$ then $\large \sum_{i=2}^{k+2}\binom{i}{2}>\frac{(k+1)^3}{6}$

17. helder_edwin

so $\large \sum_{i= 2}^{k+2}=\sum_{i=2}^{k+1}\binom{i}{2}+\binom{k+2}{2} >\frac{k^3}{6}+\frac{(k+2)!}{2!k!}$

18. helder_edwin

$\large =\frac{k!k^3+3(k+2)!}{6k!}$

19. helder_edwin

$\large =\frac{k^3+3(k+2)(k+1)(k-1)!}{6}$

20. anonymous

How did you do that last simplification?

21. helder_edwin

$\large =\frac{k^3+3(k^2+3k+2)(k-1)!}{6}$

22. anonymous

Dividing 3(k+2)! with k!

23. helder_edwin

i factored k! from the numerator

24. helder_edwin

remember this $\large n!=n(n-1)!$

25. helder_edwin

26. helder_edwin

it should be $\large =\frac{k![k^3+3(k+2)(k+1)]}{6k!}=\frac{k^3+3(k+2)(k+1)}{6}$

27. anonymous

Oh now I see it!

28. anonymous

Thank you so much, really appreciate it :D

29. helder_edwin

$\large =\frac{k^3+3k^2+9k+6}{6}$

30. anonymous

Obviously i didn't still don't get the part where you factor the k! out from 3(k+2)!

31. helder_edwin

$\large \frac{(k^3+3k^2+3k+1)+6k+5}{6}$

32. helder_edwin

using the (recursive) definition of factorial $\large (k+2)!=(k+2)(k+1)!=(k+2)(k+1)k!$

33. helder_edwin

now resuming: since k>1 $\large =\frac{(k+1)^3+6k+5}{6}>\frac{(k+1)^3}{6}$

34. helder_edwin

q.e.d.

35. anonymous

Oh so you're just saying that just as 2! = 2*1 the number before k+2 must have been k+1, right?

36. anonymous

110% got it! Wonderful! Once again, thank you!

37. helder_edwin

u r welcome