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 January 6th, 2011, 02:42 PM #1 Member   Joined: Jun 2010 Posts: 38 Thanks: 0 Binomial Theorem question Find the coefficient of $u^1^1$ in the expansion of $(3+u)^7 (5-2u)^6$
 January 6th, 2011, 03:08 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Binomial Theorem question First, use the binomial theorem to obtain the first three terms of each expansion: $(3+u)^7=u^7+21u^6+189u^5+\cdots$ $(5-2u)^6=64u^6-960u^5+6000u^4+\cdots$ Now, find the sum of the products of terms from each expansion where the sum of the exponents is 11: $u^7\cdot6000u^4-21u^6\cdot960u^5+189u^5\cdot64u^6$ Factor: $u^{11}$$6000-21\cdot960+189\cdot64$$$ Combine within the parentheses: $-2064u^{11}$

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