BdMO National Junior 2016/1
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
BdMO National Junior 2016/1
Find the value of the exprssion in the adjacent diagram.
Re: BdMO National Junior 2016/1
We can write the expression in this form:
$1+......(1+2(1+2(1+2)))...)$ where the number of brackets is $2016$. [We will call "Brackets" a pair of starting and ending brackets]
If we take the whole expression under brackets we will get the number $2017$.
If we encounter the first brackets, the result = $(1+2)=3$
If we encounter the second brackets, the result = $(1+2 \times 3)=7$
If we encounter the third brackets, the result = $(1+2 \times 7)=15$
In this way, if we encounter the $2017^{th}$ brackets, we can find the value of the expression.
We get a sequence: $3,7,15,31,...$ where the difference increases through: $4,8,16,32,...$ This sequence has $2016$ terms, easy to prove.
So the value of the expression is the sum of the total deference and the first term (3).
Total deference: $\frac{a(r^n1)}{r1}=\frac{4(2^{2016}1)}{21}=4(2^{2016}1)$
The value of the expression=$4(2^{2016}1)+3$
[There may be an easy method, this is what I did in the exam hall]
$1+......(1+2(1+2(1+2)))...)$ where the number of brackets is $2016$. [We will call "Brackets" a pair of starting and ending brackets]
If we take the whole expression under brackets we will get the number $2017$.
If we encounter the first brackets, the result = $(1+2)=3$
If we encounter the second brackets, the result = $(1+2 \times 3)=7$
If we encounter the third brackets, the result = $(1+2 \times 7)=15$
In this way, if we encounter the $2017^{th}$ brackets, we can find the value of the expression.
We get a sequence: $3,7,15,31,...$ where the difference increases through: $4,8,16,32,...$ This sequence has $2016$ terms, easy to prove.
So the value of the expression is the sum of the total deference and the first term (3).
Total deference: $\frac{a(r^n1)}{r1}=\frac{4(2^{2016}1)}{21}=4(2^{2016}1)$
The value of the expression=$4(2^{2016}1)+3$
[There may be an easy method, this is what I did in the exam hall]
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: BdMO National Junior 2016/1
Thanks for the solution.It was not good for me to not get the answer of the Junior 1st question.I didn't able to solve it.Tasnood wrote: ↑Fri Feb 16, 2018 7:42 pmWe can write the expression in this form:
$1+......(1+2(1+2(1+2)))...)$ where the number of brackets is $2016$. [We will call "Brackets" a pair of starting and ending brackets]
If we take the whole expression under brackets we will get the number $2017$.
If we encounter the first brackets, the result = $(1+2)=3$
If we encounter the second brackets, the result = $(1+2 \times 3)=7$
If we encounter the third brackets, the result = $(1+2 \times 7)=15$
In this way, if we encounter the $2017^{th}$ brackets, we can find the value of the expression.
We get a sequence: $3,7,15,31,...$ where the difference increases through: $4,8,16,32,...$ This sequence has $2016$ terms, easy to prove.
So the value of the expression is the sum of the total deference and the first term (3).
Total deference: $\frac{a(r^n1)}{r1}=\frac{4(2^{2016}1)}{21}=4(2^{2016}1)$
The value of the expression=$4(2^{2016}1)+3$
[There may be an easy method, this is what I did in the exam hall]

 Posts: 17
 Joined: Thu Feb 01, 2018 11:28 am
 Location: Sylhet
Re: BdMO National Junior 2016/1
the answer is worng i am 80% sure
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: BdMO National Junior 2016/1
What is the wrong step?

 Posts: 17
 Joined: Thu Feb 01, 2018 11:28 am
 Location: Sylhet
Re: BdMO National Junior 2016/1
we have to start from the middle bracket. Then we get 3=2^21.Then we have multiplicate 2 and +1. we get 2^n1*2+1 =2^n+12+1=2^n+11.So when there is 1 bracket we get 7[1+2(1+2)] =2^31. So when there is 2016 bracket we get 2^20181. It is the desired result.
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: BdMO National Junior 2016/1
prottoy das wrote: ↑Tue Feb 20, 2018 9:54 pmwe have to start from the middle bracket. Then we get 3=2^21.Then we have multiplicate 2 and +1. we get 2^n1*2+1 =2^n+12+1=2^n+11.So when there is 1 bracket we get 7[1+2(1+2)] =2^31. So when there is 2016 bracket we get 2^20181. It is the desired result.
You can know about latex here.
http://matholympiad.org.bd/forum/viewto ... =25&t=2#p7
 samiul_samin
 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: BdMO National Junior 2016/1
samiul_samin wrote: ↑Fri Feb 16, 2018 6:37 pmFind the value of the exprssion in the adjacent diagram.Screenshot_201802161832571.png