## BdMO National Junior 2016/1

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
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.
Screenshot_2018-02-16-18-32-57-1.png

Tasnood
Posts: 73
Joined: Tue Jan 06, 2015 1:46 pm

### 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: $\frac{a(r^n-1)}{r-1}=\frac{4(2^{2016}-1)}{2-1}=4(2^{2016}-1)$
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

Tasnood wrote:
Fri Feb 16, 2018 7:42 pm
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: $\frac{a(r^n-1)}{r-1}=\frac{4(2^{2016}-1)}{2-1}=4(2^{2016}-1)$
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]
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.

prottoy das
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

prottoy das wrote:
Tue Feb 20, 2018 9:47 pm
the answer is worng i am 80% sure
What is the wrong step?

prottoy das
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=2^2-1.Then we have multiplicate 2 and +1. we get 2^n-1*2+1 =2^n+1-2+1=2^n+1-1.So when there is 1 bracket we get 7[1+2(1+2)] =2^3-1. So when there is 2016 bracket we get 2^2018-1. 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 pm
we have to start from the middle bracket. Then we get 3=2^2-1.Then we have multiplicate 2 and +1. we get 2^n-1*2+1 =2^n+1-2+1=2^n+1-1.So when there is 1 bracket we get 7[1+2(1+2)] =2^3-1. So when there is 2016 bracket we get 2^2018-1. It is the desired result.

You can know about latex here.