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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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