For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What

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For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What [#permalink]

08 Mar 2012, 15:48

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For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What is the value of <3>*<2>?

(A) 60
(B) 116
(C) 210
(D) 263
(E) 478

Correct answer is 210.

I can not understand why we put 3 in the formula above 6 times and 2 only 4?

Thank you so much guys.
Serge.

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Re: For All Integers [#permalink]

08 Mar 2012, 15:56

For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What is the value of <3>*<2>?
(A) 60
(B) 116
(C) 210
(D) 263
(E) 478

Since <x>=2x+(2x-1)+(2x-2)+......2+1=1+2+..+(2x-2)+(2x-1)+2x then <x> is basically the sum of all integers from 1 to 2x, inclusive.

Hence <3> is the sum of all integers from 1 to 2*3=6 and <2> is the sum of all integers from 1 to 2*2=4 --> <3>=21 and <2>=10 --> <3>*<2>=21*10=210.

Answer: C.

Hope it's clear.
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Re: For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What [#permalink]

08 Mar 2012, 16:01

Thank you SO much, Bunuel.

Now it is clear!!!!

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Re: For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What [#permalink]

07 May 2012, 11:57

SergeNew wrote:

The problem can be understand very simply it is stating that for any value as <x> ,x=2x+2x-1+2x-2 this process is continued till we get 2+1

so the question want us to calculate <3>*<2>

so lets substitute 3 in place of x then we get <3>=2(3)+(2*3-1)+(2*3-2)+(2*3-3)+(2*3-4)+(2*3-5) we can stop here because if we go further we will get zero and negative values to prove lets 2*3-6=6-6=0 so 0 and the numbers will be decreasing by 1 as the procedure goes , so as question states we should stop when we reach 2+1

so <3>=2(3)+(2*3-1)+(2*3-2)+(2*3-3)+(2*3-4)+(2*3-5) => <3> = 6+(6-1)+(6-2)+(6-3)+(6-4)+(6-5)=> 6+5+4+3+2+1=> 21 so for <3> we got 21 and continue the same procedure for <2>

<2>=2(2)+(2*2-1)+(2*2-2)+(2*2-3) we should stop here because we should not go beyond 1 as question states
<2>=>4+(4-1)+(4-2)+(4-3)=>4+3+2+1=10

so <3>*<2>=21*10=210 i.e c

Hope it's concise and clear.

Re: For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What [#permalink] 07 May 2012, 11:57

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For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What

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For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What

For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What

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