solve [2+1][2^2+1][2^4+1][2^8+1][2^16+1]...........[2^256+1]

M4maths Previous Puzzles

17 Aug 2012 series of 2

solve
[2+1][2^2+1][2^4+1][2^8+1][2^16+1]...........[2^256+1]

OPtion

1) 19865432443242
2) 2^256+2^128+1
3) 2^512-1
4) 2^512+2^128+1
5) 2^512-2^256+2^128-1
6) 2^512+1
7) 2^511
8) 2^511-1
9) 2^511+1
10) none of these

Solution

[2+1][2^2+1][2^4+1][2^8+1][2^16+1]...........[2^256+1]
on multiplying [2-1] in the numerator and denominator, we get
[2-1][2+1][2^2+1][2^4+1][2^8+1][2^16+1]...........[2^256+1]/[2-1]
now we know that (a+b)(a-b) = a^2-b^2 so next line will be
[2^2-1][2^2+1][2^4+1][2^8+1][2^16+1]...........[2^256+1]/1
=[2^4-1][2^4+1][2^8+1][2^16+1]...........[2^256+1]
=[2^8-1][2^8+1][2^16+1]...........[2^256+1]
=[2^16-1][2^16+1]...........[2^256+1]
=[2^32-1][2^32+1]...........[2^256+1]
solve this, final result will be 2^512-1

Correct Option: 3 Best Solution (5)

Dr.Dipin Singh   12 years AGO

[2+1][2^2+1][2^4+1][2^8+1][2^16+1]...........[2^256+1] = [2^2-1][2^2+1][2^4+1][2^8+1][2^16+1]...........[2^256+1] = [2^4-1][2^4+1][2^8+1][2^16+1]...........[2^256+1] =
[2^256-1]*[2^256+1] = 2^512 -1 ... option 3

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Surbhi Sharma   12 years AGO

[2+1] can be written as=[2^2-1]
[2+1][2^2+1] can be written as=[2^4-1]
thus
[2+1][2^2+1][2^4+1][2^8+1][2^16+1]....[2^256+1]
can be written as=[2^512-1]

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Jason   12 years AGO

[2+1][2^2+1][2^4+1][2^8+1][2^16+1]...........[2^256+1] = [2^256-1]*[2^256+1]= 2^512 -1

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ajitha   12 years AGO

ans:2^512-1
[2+1][2^2+1][2^4+1][2^8+1][2^16+1]...........[2^256+1]


2+1=3 written as 2^2-1
[2+1][2^2+1]=15 written as 2^4-1
[2+1][2^2+1][2^4+1]=255 written as 2^8-1
[2+1][2^2+1][2^4+1][2^8+1]=65535 written as 2^16-1
similar way we can write the value of entire above series as 2^512-1

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birenparekh   12 years AGO

il is geometric progressoin..
sum of all is 1+2^1+2^3+2^4+.....+2^511
so a=1,r=2/1=2 and n=512(total no of term)
eq sum=a(r^n-1)/(r-1)
solving it..sum=1(2^512-1)/(2-1)
=2^512-1//option (3)

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