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# 'BODMAS' Rule

1+2(1+2(1+2(1+...........)))))))))))).........))) If the number of brasckets are 2016 , then what is the value of this problem?

Note by Shithil Islam
1 year, 9 months ago

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The generel form would be -( 2^(n+2) ) -1 where n is the number of brackets

- 1 year, 6 months ago

I would rewrite this as a sequence, where $$a_{1}=1+2, a_{2}=1+2(1+2)$$ and so on. finding the values of each term would give you 3,7,15,31... and this ends up being $$a_{n}=2^{n+1} -1$$. Therefore, the value you are finding for this problem would be $$2^{2018}-1$$. Hope this is correct and if it's not, tell me and I will revise it :D

- 1 year, 8 months ago

Oh, now I get it. You are right :) :)

- 1 year, 8 months ago

Thanks and I am so glad that you got it!

- 1 year, 8 months ago

2^2018-1

- 1 year, 9 months ago

How can i find it? Please describe me

- 1 year, 9 months ago