Stitz-Zeager_College_Algebra_e-book

# 5 2 since 225 is negative we start at the positive x

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Unformatted text preview: g the summation using ‘n’ as the index. 566 Sequences and the Binomial Theorem If we rewrite the quantity 2a + (n − 1)d as a + (a + (n − 1)d) = a1 + an , we get the formula S=n a1 + an 2 A helpful way to remember this last formula is to recognize that we have expressed the sum as the product of the number of terms n and the average of the ﬁrst and nth terms. To derive the formula for the geometric sum, we start with a geometric sequence ak = ark−1 , k ≥ 1, and let S once again denote the sum of the ﬁrst n terms. Comparing S and rS , we get S = a + ar + ar2 + . . . + arn−2 + arn−1 rS = ar + ar2 + . . . + arn−2 + arn−1 + arn Subtracting the second equation from the ﬁrst forces all of the terms except a and arn to cancel out and we get S − rS = a − arn . Factoring, we get S (1 − r) = a (1 − rn ). Assuming r = 1, we can divide both sides by the quantity (1 − r) to obtain S=a 1 − rn 1−r If we distribute a through the numerator, we get a − arn = a1 − an+1 which yields the formula S= a1 − an+1...
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