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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online