Question

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...

Answer 1

Have you ever learned mathematical induction?

Answer 2

Mathematical induction will do..

Answer 3

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

Answer 4

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

Answer 5

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?

Answer 6

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

Answer 7

Yes I do

Answer 8

Nothing different..choose k..

Answer 9

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

Answer 10

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

Answer 11

Yes..: )

Answer 12

A little \(\LaTeX\) tip

you can use {n\choose r} for \[{n\choose r}\]

Answer 13

\[\binom{n}{r}\]

or \binom{n}{r}

Answer 14

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

Answer 15

\[ {2\choose 2} + {3\choose 2}+ {4\choose 2}+...++ {k+2\choose 2} > \frac{ (k+1)^{3} }{ 6}\]


\[\frac{ k ^{3} }{ 6 }+{k+2\choose 2} = \frac{ k ^{3} }{ 6 } + \frac{ (k+2)! }{ 2!k!}=\frac{ 2!k!k ^{3}+6(k+2)! }{ 2!k!6 }=\frac{ k!k ^{3}+3(k+2)! }{ k!6 }\]

I don't know how to simplify the resy, could someone please help me?

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

How did you do that last simplification?

Answer 21

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

Answer 22

Dividing 3(k+2)! with k!

Answer 23

i factored k! from the numerator

Answer 24

remember this
\[ \large n!=n(n-1)! \]

Answer 25

i made a mistake

Answer 26

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

Answer 27

Oh now I see it!

Answer 28

Thank you so much, really appreciate it :D

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

q.e.d.

Answer 35

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

Answer 36

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

Answer 37

u r welcome