Question

help wid part b
http://prntscr.com/3isdul

Answer 1

\[1^2 + 2^2 + 3^2 \cdots +n^2\]\[1 +\left( 2\binom{2}{2} + 2\right) + \left(2\binom{3}{2} + 3\right) + \cdots + \left( 2\binom{n}{2}+n\right) \]Just a thought. You can group those combination terms and use that identity in \(\rm (a)\).

Answer 2

thats a damn good start i think :

\[\left[ 1 +2+3+...+n\right] + 2\left[ \binom{2}{2} + \binom{3}{2} + \cdots + \binom{n}{2}\right]\]

Answer 3

left sum requires the n(n+1)/2 formula, but it should be okay to use :

\[\left[ \dfrac{n(n+1)}{2}\right] + 2\left[ \binom{n+1}{3}\right]\]

Answer 4

just need to simplify :) thank you xD

Answer 5

That's the MathOverflow thread right? :P

I also like the \(\binom{n}{2}\) proof there.

Answer 6

while googling for the solution to present problem, i got that interesting proof using triangles ^

Answer 7

OMG ! you knew it before ha

Answer 8

yeah they all are magic of pascal triangle

Answer 9

you talking about this :
\[1 + 2+3+\cdots n = \binom{n}{2}\]

right ?

Answer 10

Yes.

Answer 11

in general :
\[\sum \limits_{k=r}^n \binom{k}{r} = \binom{n+1}{r}\]

Answer 12

Interesting result.

Answer 13

plugging r = 1 gives \(\binom{n}{2}\) proof

Answer 14

indeed it is very interesting, it has a name : hockeystick rule
http://www.learner.org/courses/mathilluminated/images/units/2/hockeysticks.gif

Answer 15

Thanks for sharing this. :)

I haven't really ever thought about combinatorics.

Answer 16

*corrected
\[\sum \limits_{k=r}^n \binom{k}{r} = \binom{n+1}{r+1} \]

pascal's triangle is a nice drill for combinatorics...

Answer 17

Just a thought, then... can we use\[1 + 2 + 3 + \cdots + n = \binom{n}{2}\]in this problem instead of \(n(n+1)/2\), using any of those identities?

Answer 18

\[\binom{n+1}{3}\]and\[\binom{n}{2}\]are related in the same way.

Answer 19

oh have to think...

*corrected
\[1 + 2 + 3 + \cdots + n = \binom{n+1}{2} \]

Answer 20

already we're using \(\binom{n+1}{3}\)

you want to avoid using that n(n+1)/2 formula directly/blindly is it

Answer 21

Oh my... I confused \(2^n\) with that. So sorry.

Answer 22

2^n is for rows
hockey stick is for diagonals

Answer 23

Yuppers, my brain sucked for a minute.

Answer 24

haha im stuck on this pascal triangle unit from last 6 days lol

Answer 25

here is the modified proof using pascal triangle in its fullest glory :

Answer 26

using :
\[m^2 = 2\binom{m}{2} + \binom{m}{1}\]

Answer 27

:P

Answer 28

\[\binom{1}{1} +\left( 2\binom{2}{2} + \binom{2}{1}\right) + \left(2\binom{3}{2} + \binom{3}{1}\right) + \cdots + \left( 2\binom{n}{2}+\binom{n}{1}\right)\]

\[
\left[\binom{1}{1} + \binom{2}{1} + \cdots \binom{n}{1}\right] + 2\left[ \binom{2}{2} + \binom{3}{2} + \cdots + \binom{n}{2}\right]

\]

Answer 29

it uses only hockeystick rule now, thanks for hinting that :)

Answer 30

I have no idea what you're thanking me for, but I'd just go with the flow and say you're welcome. :P

Answer 31

thanking for that^^

Answer 32

Oh, OK :)