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ParthKohli
 one year ago
As promised. Many of you may have solved this, but anyway.
ParthKohli
 one year ago
As promised. Many of you may have solved this, but anyway.

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2A number sequence \(a_1, a_2, \cdots, a_n\) is such that\[a_1 =0\]\[a_2  = a_1 + 1\]\[\vdots\]\[a_n = a_{n1} + 1\]Prove that the arithmeticmean of \(a_1, a_2, \cdots, a_n \) is \(\ge 1/2\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2@imqwerty @mathmath333 @perl

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2don't we simply get a sequence of whole numbers ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\(a_2\) can also be 1, for instance.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2A hint is that you don't want the absolute value there. What do you do?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Oh parth you give hints too early

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Sorry, I'm an impatient guy. :

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\sum\limits_{i=1}^{n} {a_n}^2~~=~~\sum\limits_{i=1}^{n} (a_{n1}+1)^2\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2should i try telescoping ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2You were at least able to extract meaning out of that hint.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\sum\limits_{i=1}^{n} {a_n}^2{a_{n1}}^2~~=~~\sum\limits_{i=1}^{n} 2a_{n1}+1\] ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2idk what to do next but if i had pen and paper wid me i would try and see if it telescopes..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, things cancel out.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2The rightside is very important.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2let me fix the typoes in indices \[\sum\limits_{i=\color{Red}{2}}^{n} {a_\color{Red}{i}}^2{a_{\color{Red}{i}1}}^2~~=~~\sum\limits_{i=\color{red}{2}}^{n} 2a_{\color{red}{i}1}+1\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, it's mostly about the RHS now.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I think i need to zero the index on right hand side

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Make both the upper limits \(n+1\).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[{a_n}^2{a_1}^2~~=~~(n1)2a_n+2\sum\limits_{i=\color{red}{1}}^{n} a_{\color{red}{i}} \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2looks completing the square will do the job

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, you can also use \(a_1 = 0\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[({a_n}+1)^2~~=~~{a_1}^2+n+2\sum\limits_{i=\color{red}{1}}^{n} a_{\color{red}{i}}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[0~~\le ~~n+2\sum\limits_{i=\color{red}{1}}^{n} a_{\color{red}{i}}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Great, well done :)
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