Question

As promised. Many of you may have solved this, but anyway.

Answer 1

A number sequence \(a_1, a_2, \cdots, a_n\) is such that\[a_1 =0\]\[|a_2 | = |a_1 + 1|\]\[\vdots\]\[|a_n| = |a_{n-1} + 1|\]Prove that the arithmetic-mean of \(a_1, a_2, \cdots, a_n \) is \(\ge -1/2\).

Answer 2

@ganeshie8

Answer 3

@imqwerty @mathmath333 @perl

Answer 4

don't we simply get a sequence of whole numbers ?

Answer 5

Yes, of course.

Answer 6

Of integers.

Answer 7

\(a_2\) can also be -1, for instance.

Answer 8

Oh I see it.

Answer 9

A hint is that you don't want the absolute value there. What do you do?

Answer 10

Oh parth you give hints too early

Answer 11

Sorry, I'm an impatient guy. :|

Answer 12

\[\sum\limits_{i=1}^{n} {a_n}^2~~=~~\sum\limits_{i=1}^{n} (a_{n-1}+1)^2\]

Answer 13

:D

Answer 14

should i try telescoping ?

Answer 15

You were at least able to extract meaning out of that hint.

Answer 16

Nah, just expand.

Answer 17

\[\sum\limits_{i=1}^{n} {a_n}^2-{a_{n-1}}^2~~=~~\sum\limits_{i=1}^{n} 2a_{n-1}+1\]

?

Answer 18

idk what to do next but if i had pen and paper wid me i would try and see if it telescopes..

Answer 19

Yeah, things cancel out.

Answer 20

The right-side is very important.

Answer 21

let me fix the typoes in indices

\[\sum\limits_{i=\color{Red}{2}}^{n} {a_\color{Red}{i}}^2-{a_{\color{Red}{i}-1}}^2~~=~~\sum\limits_{i=\color{red}{2}}^{n} 2a_{\color{red}{i}-1}+1\]

Answer 22

Yeah, it's mostly about the RHS now.

Answer 23

I think i need to zero the index on right hand side

Answer 24

Make both the upper limits \(n+1\).

Answer 25

\[{a_n}^2-{a_1}^2~~=~~(n-1)-2a_n+2\sum\limits_{i=\color{red}{1}}^{n} a_{\color{red}{i}} \]

Answer 26

looks completing the square will do the job

Answer 27

Yeah, you can also use \(a_1 = 0\)

Answer 28

\[({a_n}+1)^2~~=~~{a_1}^2+n+2\sum\limits_{i=\color{red}{1}}^{n} a_{\color{red}{i}}\]

Answer 29

\[0~~\le ~~n+2\sum\limits_{i=\color{red}{1}}^{n} a_{\color{red}{i}}\]

Answer 30

Yaay!

Answer 31

Great, well done :)

Answer 32

ty ty :)