Question

Just two questions from a math contest I participated, I solved it, but can you? :P

Simplify this:
http://puu.sh/q0hv
and this:
http://puu.sh/q0i9

P.S.: The purpose of this question is just for this: http://puu.sh/q0il :P

Have fun! ;D

Answer 1

@joemath314159 Right answer :D

Answer 2

First one can be changed from\[{1 \over \sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\]This leaves us with the series\[1+(\sqrt{2}-1)+(\sqrt{3}-\sqrt2)+...+(\sqrt{2012}-\sqrt{2011})\]Which is a telescoping series, so it can be simplified to just \[\sqrt{2012}\]

Answer 3

As for the second one, it can be simplified to \[{(n+1)(2n-1) \over (n-1)(2n+1)}\]If we start multiplying things together, we get \[{(n+1)(2n-1) \over (n-1)(2n+1)}\cdot {(n+2)(2n+1) \over (n)(2n+3)}\cdot{(n+3)(2n+3) \over (n+1)(2n+5)}\]

Answer 4

This is another telescoping series, so our answer can be simplified to \[{100! \over 98!}\cdot{3 \over 197}\]I think...

Answer 5

Nope ;(

Answer 6

i think its 4950/67, whatever that is lol

Answer 7

I didn't do it this way, I took what I got from f(2) to f(6) and observed the pattern, and give 5050/67

Answer 8

Well For the first one the last and first terms in consecutive conjugates cancel out.

Answer 9

Here's how I proceed: http://puu.sh/q0yE

Answer 10

my bad, forgot one last term in the sequence, its 5050/67

Answer 11

I figured out where my telescoping series was going wrong. If it goes to 100, it can be simplified to \[{(n+98)(n+99) \over (n-1)n}\cdot {2n-1 \over 2n+197}\]Where \(n=2\). Thus, the sume would be \[{100\cdot101 \over 2}\cdot{3 \over 201}={5050\over67}\]

Answer 12

You people are so cool x]

Both questions are solved! Hope you guys had fun ;D

Answer 13

Definitely did. Thanks for the problems. :)