BdMO National Higher Secondary 2008/1

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Moon
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BdMO National Higher Secondary 2008/1

Unread post by Moon » Sun Feb 06, 2011 11:17 pm

Problem 1:
The Sum of the first $2008$ odd positive integer is subtracted from the sum of the first $2008$ even positive integers. Find the result.
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Tasnood
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Re: BdMO National Higher Secondary 2008/1

Unread post by Tasnood » Fri Feb 09, 2018 8:53 pm

Sum of the first $2008$ even positive integers=$2+4+6+...+n$
Sum of the first $2008$ odd positive integers=$1+3+5+7+...+n$ [Where $n$ denotes the $2008$-th term of the sequence]


Difference between the first terms of two sequence=$1$
Difference between the second terms of two sequence=$1$
So,Difference between the $n$-th terms of two sequence=$1$

So, the result=$1 \times 2008$=$2008$
Last edited by Tasnood on Fri Feb 09, 2018 9:58 pm, edited 2 times in total.

aritra barua
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Re: BdMO National Higher Secondary 2008/1

Unread post by aritra barua » Fri Feb 09, 2018 9:55 pm

Tasnood,$0$ is not a positive integer. Now,the sum of the first $n$ even numbers is $n(n+1)$ and for the case of odd numbers,that is $n^2$.So,by the question, the answer is $2008(2008+1)-2008^2$=$2008$.

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