Waste less time on Facebook — follow Brilliant.
×

'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
7 months, 2 weeks ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

The generel form would be -( 2^(n+2) ) -1 where n is the number of brackets Aditya Kumar · 4 months, 2 weeks ago

Log in to reply

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 Margaret Zheng · 7 months ago

Log in to reply

@Margaret Zheng Oh, now I get it. You are right :) :) Shithil Islam · 7 months ago

Log in to reply

@Shithil Islam Thanks and I am so glad that you got it! Margaret Zheng · 7 months ago

Log in to reply

2^2018-1 Asif Mujawar · 7 months, 2 weeks ago

Log in to reply

@Asif Mujawar How can i find it? Please describe me Shithil Islam · 7 months, 2 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...