Stitz-Zeager_College_Algebra_e-book

# 11 it is worth noting that all of the pathologies of

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Unformatted text preview: ± n=m p c an = c n=m an , for any real number c. n=m p • j an = n=m p an + n=m p • an , for any natural number m ≤ j < j + 1 ≤ p. n=j +1 p+r an = n=m bn n=m an−r , for any whole number r. n=m+r We now turn our attention to the sums involving arithmetic and geometric sequences. Given an arithmetic sequence ak = a + (k − 1)d for k ≥ 1, we let S denote the sum of the ﬁrst n terms. To derive a formula for S , we write it out in two diﬀerent ways S= a + (a + d) + . . . + (a + (n − 2)d) + (a + (n − 1)d) S = (a + (n − 1)d) + (a + (n − 2)d) + . . . + (a + d) + a If we add these two equations and combine the terms which are aligned vertically, we get 2S = (2a + (n − 1)d) + (2a + (n − 1)d) + . . . + (2a + (n − 1)d) + (2a + (n − 1)d) The right hand side of this equation contains n terms, all of which are equal to (2a + (n − 1)d) so we get 2S = n(2a + (n − 1)d). Dividing both sides of this equation by 2, we obtain the formula S= 3 n (2a + (n − 1)d) 2 To see why, try writin...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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