ParthKohli
  • ParthKohli
As promised. Many of you may have solved this, but anyway.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • ParthKohli
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 \(\ge -1/2\).
ParthKohli
  • ParthKohli
@ganeshie8
ParthKohli
  • ParthKohli
@imqwerty @mathmath333 @perl

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ganeshie8
  • ganeshie8
don't we simply get a sequence of whole numbers ?
ParthKohli
  • ParthKohli
Yes, of course.
ParthKohli
  • ParthKohli
Of integers.
ParthKohli
  • ParthKohli
\(a_2\) can also be -1, for instance.
ganeshie8
  • ganeshie8
Oh I see it.
ParthKohli
  • ParthKohli
A hint is that you don't want the absolute value there. What do you do?
ganeshie8
  • ganeshie8
Oh parth you give hints too early
ParthKohli
  • ParthKohli
Sorry, I'm an impatient guy. :|
ganeshie8
  • ganeshie8
\[\sum\limits_{i=1}^{n} {a_n}^2~~=~~\sum\limits_{i=1}^{n} (a_{n-1}+1)^2\]
ParthKohli
  • ParthKohli
:D
ganeshie8
  • ganeshie8
should i try telescoping ?
ParthKohli
  • ParthKohli
You were at least able to extract meaning out of that hint.
ParthKohli
  • ParthKohli
Nah, just expand.
ganeshie8
  • ganeshie8
\[\sum\limits_{i=1}^{n} {a_n}^2-{a_{n-1}}^2~~=~~\sum\limits_{i=1}^{n} 2a_{n-1}+1\] ?
ganeshie8
  • ganeshie8
idk what to do next but if i had pen and paper wid me i would try and see if it telescopes..
ParthKohli
  • ParthKohli
Yeah, things cancel out.
ParthKohli
  • ParthKohli
The right-side is very important.
ganeshie8
  • ganeshie8
let me fix the typoes in indices \[\sum\limits_{i=\color{Red}{2}}^{n} {a_\color{Red}{i}}^2-{a_{\color{Red}{i}-1}}^2~~=~~\sum\limits_{i=\color{red}{2}}^{n} 2a_{\color{red}{i}-1}+1\]
ParthKohli
  • ParthKohli
Yeah, it's mostly about the RHS now.
ganeshie8
  • ganeshie8
I think i need to zero the index on right hand side
ParthKohli
  • ParthKohli
Make both the upper limits \(n+1\).
ganeshie8
  • ganeshie8
\[{a_n}^2-{a_1}^2~~=~~(n-1)-2a_n+2\sum\limits_{i=\color{red}{1}}^{n} a_{\color{red}{i}} \]
ganeshie8
  • ganeshie8
looks completing the square will do the job
ParthKohli
  • ParthKohli
Yeah, you can also use \(a_1 = 0\)
ganeshie8
  • ganeshie8
\[({a_n}+1)^2~~=~~{a_1}^2+n+2\sum\limits_{i=\color{red}{1}}^{n} a_{\color{red}{i}}\]
ganeshie8
  • ganeshie8
\[0~~\le ~~n+2\sum\limits_{i=\color{red}{1}}^{n} a_{\color{red}{i}}\]
ganeshie8
  • ganeshie8
Yaay!
ParthKohli
  • ParthKohli
Great, well done :)
ganeshie8
  • ganeshie8
ty ty :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.