\[1-2+3-4+\cdot \cdot \cdot+(-1)^{p+1} \cdot p \geq 2012\]
find the samllest value of $p$
[i'm not sure of the problem number
]Mymemensingh 2012 Secondary 07
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nafistiham
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if
\[1-2+3-4+\cdot \cdot \cdot+(-1)^{p+1} \cdot p \geq 2012\]
find the samllest value of $p$
[i'm not sure of the problem number ]
\[1-2+3-4+\cdot \cdot \cdot+(-1)^{p+1} \cdot p \geq 2012\]
find the samllest value of $p$
[i'm not sure of the problem number
]\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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sakib.creza
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Re: Mymemensingh 2012 Secondary 07
1-2+3-4+5-6....
-1+-1+-1+......
for up to nth term of the sequence, where n is even, the summation of the terms is (n/2)(-1), which is negative.
the next term is odd, (n+1)(-1)^(n+2), which is positive.
According to the question, (n/2)(-1)+(n+1)(-1)^(n+2) must be greater than or equal to 2012.
So least value of n occurs when the sum is equal to 2012.
as n+2 is positive, we can write, (-n/2)+(n+1)=2012.
So, n=4022.
So, p=4023.
-1+-1+-1+......
for up to nth term of the sequence, where n is even, the summation of the terms is (n/2)(-1), which is negative.
the next term is odd, (n+1)(-1)^(n+2), which is positive.
According to the question, (n/2)(-1)+(n+1)(-1)^(n+2) must be greater than or equal to 2012.
So least value of n occurs when the sum is equal to 2012.
as n+2 is positive, we can write, (-n/2)+(n+1)=2012.
So, n=4022.
So, p=4023.