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Source: — Problem Solving |
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by cramya » Sun Oct 05, 2008 5:39 pm
The answer should be A)

X1 = 3 (2^1+1)
X2 = 5 (2^2+1)
X3 = 9 (2^3+1)

....

THEREFORE X20 = 2 ^ 20 + 1
X19 = 2^19+1

X20-X19
= 2^20+1 - (2^19+1)
= 2^20 + 1 - 2^19 - 1
= 2 ^ 20 - 2 ^ 19

TAKING COMMON FACTOR 2^19 OUT
2^19(2 ^ 1 - 1)
= 2 ^ 19
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by cramya » Sun Oct 05, 2008 5:41 pm
Just saw the OA and its A)
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by cramya » Sun Oct 05, 2008 5:43 pm
My thought is on these type of problems find a pattern for the terms in the sequence as calculating it manually would take more than 2 minutes to solve.

For eg: In the problem above expressing each term in terms of 2 ^ something
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by 4meonly » Mon Oct 06, 2008 9:27 am
cramya wrote:The answer should be A)

X1 = 3 (2^1+1)
X2 = 5 (2^2+1)
X3 = 9 (2^3+1)
I dont understand how you made it from the question without any powers?
I question is not correct, could you please post in correct manner?
thanx!
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by cramya » Mon Oct 06, 2008 2:11 pm
Hi 4meonly,

X1 = 3
X2 = 5
X3 = 9

I was saying that the first term X1 which is 3 can be expressed as 2 ^1 + 1
2nd term x2 which is 5 can be expressed as 2 ^2+1

Similarly
x19 can be expressed as 2 ^ 19 + 1
x20 can be expressed as 2 ^ 20 +1

Hope this helps!
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by sudhir3127 » Sun Nov 30, 2008 12:15 am
cramya wrote:The answer should be A)

X1 = 3 (2^1+1)
X2 = 5 (2^2+1)
X3 = 9 (2^3+1)

....

THEREFORE X20 = 2 ^ 20 + 1
X19 = 2^19+1

X20-X19
= 2^20+1 - (2^19+1)
= 2^20 + 1 - 2^19 - 1
= 2 ^ 20 - 2 ^ 19

TAKING COMMON FACTOR 2^19 OUT
2^19(2 ^ 1 - 1)
= 2 ^ 19
Nice way of solving ... but if u wanna do it real quick u can cut down on most of the steps...

X1 = 3
X2 = 5
X3 = 9

now X2- X1 = 2(2^1)
X3- X1 = 4(2^2)

2,4,8,..............

thus X20 - X 19 = 2^19.
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