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Stitz-Zeager_College_Algebra_e-book

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

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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...
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