Stitz-Zeager_College_Algebra_e-book

Sin 1 2 3 cos 0 solution since there is no context

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: move is to ‘kick out’ the terms which we cannot combine and rewrite the summations so that we can combine them. To that end, we note k j =0 k k+1−j j a b = ak+1 + j k j =1 k k+1−j j a b j and k j =0 k k−j j +1 ab = j k −1 k k−j j +1 ab + bk+1 j j =0 so that k k k+1−j j a b+ j (a + b)k+1 = ak+1 + j =1 k −1 j =0 k k−j j +1 ab + bk+1 j We now wish to write k j =1 k k+1−j j a b+ j k−1 j =0 k k−j j +1 ab j as a single summation. The wrinkle is that the first summation starts with j = 1, while the second starts with j = 0. Even though the sums produce terms with the same powers of a and b, they do so for different values of j . To resolve this, we need to shift the index on the second summation so that the index j starts at j = 1 instead of j = 0 and we make use of Theorem 9.1 in the process. k−1 j =0 k k−j j +1 ab = j k−1+1 k ak−(j −1) b(j −1)+1 j−1 j =0+1 k = j =1 k ak+1−j bj j−1 We can now combine our two sums using Theorem 9.1 and simplify using Theorem 9.3 k j =1 k k+1−j j a b+ j k−1 j =0 k k−j j +1 ab = j k j =1 k = j =1 k = j =1 Using...
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