### BdMO National Junior 2016/1

Posted:

**Fri Feb 16, 2018 6:37 pm**Find the value of the exprssion in the adjacent diagram.

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Posted: **Fri Feb 16, 2018 6:37 pm**

Find the value of the exprssion in the adjacent diagram.

Posted: **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]**

$1+......(1+2(1+2(1+2)))...)$ where the number of brackets is $2016$.

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

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

Posted: **Fri Feb 16, 2018 8:09 pm**

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 isthe 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]

Posted: **Tue Feb 20, 2018 9:47 pm**

the answer is worng i am 80% sure

Posted: **Tue Feb 20, 2018 9:53 pm**

What is the wrong step?

Posted: **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.

Posted: **Tue Feb 20, 2018 10:03 pm**

prottoy das wrote: ↑Tue Feb 20, 2018 9:54 pmwe 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.

http://matholympiad.org.bd/forum/viewto ... =25&t=2#p7

Posted: **Sun Feb 17, 2019 8:54 am**

samiul_samin wrote: ↑Fri Feb 16, 2018 6:37 pmFind the value of the exprssion in the adjacent diagram.Screenshot_2018-02-16-18-32-57-1.png