Problema Solution
For positive integer values of N, let N be defined as:
N=2+4+6+...+N, if N is even and
N=1+3+5+...+N, kif N is odd.
What is the value of 2009-2008?
Answer provided by our tutors
First we need to find the sums of the Arithmetic Sequences using the formula:
the sum of the first n terms of the arithmetic sequence is
Sn = ((n+1)/2)(2a + nd)
For the sequence 1, 3, 5, 7, 9, ...2009 the first term is a = 1 and common difference d = 2
an = a + n*d
a + n*d = 2009
1 + 2*n = 2009
n = 1004
a1004 = 1 + 1004*2
the sum of the first 1004 terms of the arithmetic sequence is
S1004 = (1005/2)(2*1 + 1004*2)
S1004 = 1,010,025
For the sequence 2, 4, 6, 8, ...2008 the first term is a = 2 and common difference d = 2
an = a + n*d
2008 = 2 + 2*n
n = 2006/2
n = 1003
the sum of the first 1003 terms of the arithmetic sequence is
S1003 = (1004/2)(2*2 + 1003*2)
S1003 = 1,009,020
Now we can finally calculate
(1 + 3 + 5 +7 +...+ 2009) - (2 +4 +6 + ...+ 2008) = 1,010,025 - 1,009,020 = 1005
Another more simple way of solving this problem is by using the commutative and associative law of addition
(1 + 3 + 5 +7 +...+ 2009) - (2 +4 +6 + ...+ 2008) =
= (1 - 2) + (2 - 3) + ...(2007 - 2008) + 2009 =
= (-1)*(2008/2) + 2009 = -1004 + 2009 = 1005