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Bunuel
If \(a_1\), \(a_2\), ... are integers such that \(a_1 – a_2 + a_3 – a_4 + ........ +(-1)^{n-1}\) \(a_n = n\), for \(n ≥ 1\). Then what is the value of \(a_{51} + a_{52} + ... + a_{1023}\) ?

A. -1
B. 0
C. 1
D. 2
E. 10


Are You Up For the Challenge: 700 Level Questions

Asked: If \(a_1\), \(a_2\), ... are integers such that \(a_1 – a_2 + a_3 – a_4 + ........ +(-1)^{n-1}\) \(a_n = n\), for \(n ≥ 1\). Then what is the value of \(a_{51} + a_{52} + ... + a_{1023}\) ?

\(a_1 = 1\)
\(a_1 - a_2 = 2\)
\(a_2 = a_1 - 2 = 1 - 2 = - 1\)
\(a_1 - a_2 + a_ 3 = 3\)
\(a_3 =3-2 = 1\\
[m]a_1 - a_2 + a_ 3 - a_4\) = 4[/m]
\(a_4 = -1\)
\(a_n = (-1)^n\)

\(a_{51} + a_{52} + ... + a_{1023} = 1-1+ 1- 1+ ......+1 = 1\)

IMO C
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Bunuel
If \(a_1\), \(a_2\), ... are integers such that \(a_1 – a_2 + a_3 – a_4 + ........ +(-1)^{n-1}\) \(a_n = n\), for \(n ≥ 1\). Then what is the value of \(a_{51} + a_{52} + ... + a_{1023}\) ?

A. -1
B. 0
C. 1
D. 2
E. 10


Let’s first determine the first few terms and hope we can discover a pattern for the terms:

a1 = 1,

a1 - a2 = 2 → 1 - a2 = 2 → a2 = -1

a1 - a2 + a3 = 3 → 1 - (-1) + a3 = 3 → a3 = 1

a1 - a2 + a3 - a4 = 4 → 1 - (-1) + 1 - a4 = 4 → a4 = -1

We see that the odd numbered terms are 1 and the even numbered terms are -1. Thus, the sum of a pair of consecutive terms is 0. So we have:

(a51 + a52) + (a53 + a54) + … + (a1021 + a1022) + a1023

0 + 0 + … + 0 + 1

1

Answer: C
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Can't understand how we got A1, A2, A3 in the first place

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