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Prove-that-the-coefficient-of-x-p-in-the-expansion-of-a-0-a-1-x-a-2-x-2-a-3-x-3-a-k-x-k-n-is-n-n-0-n-1-n-2-n-3-n-k-a-0-n-0-a-1-n-1-a-2-n-2-a-3-n-3-a-k-n-k-whe




Question Number 22316 by Tinkutara last updated on 15/Oct/17
Prove that the coefficient of x^p  in the  expansion of (a_0 +a_1 x+a_2 x^2 +a_3 x^3 +...+a_k x^k )^n   is Σ((n!)/(n_0 !n_1 !n_2 !n_3 !...n_k !))a_0 ^n_0  a_1 ^n_1  a_2 ^n_2  a_3 ^n_3  ...a_k ^n_k    where n_0 , n_1 , n_2 , n_3 , ..., n_k  are all non-  negative integers subject to the  conditions n_0 +n_1 +n_2 +n_3 +...+n_k =n  and n_1 +2n_2 +3n_3 +...+kn_k =p.
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{p}} \:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{expansion}\:\mathrm{of}\:\left({a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +{a}_{\mathrm{3}} {x}^{\mathrm{3}} +…+{a}_{{k}} {x}^{{k}} \right)^{{n}} \\ $$$$\mathrm{is}\:\Sigma\frac{{n}!}{{n}_{\mathrm{0}} !{n}_{\mathrm{1}} !{n}_{\mathrm{2}} !{n}_{\mathrm{3}} !…{n}_{{k}} !}{a}_{\mathrm{0}} ^{{n}_{\mathrm{0}} } {a}_{\mathrm{1}} ^{{n}_{\mathrm{1}} } {a}_{\mathrm{2}} ^{{n}_{\mathrm{2}} } {a}_{\mathrm{3}} ^{{n}_{\mathrm{3}} } …{a}_{{k}} ^{{n}_{{k}} } \\ $$$$\mathrm{where}\:{n}_{\mathrm{0}} ,\:{n}_{\mathrm{1}} ,\:{n}_{\mathrm{2}} ,\:{n}_{\mathrm{3}} ,\:…,\:{n}_{{k}} \:\mathrm{are}\:\mathrm{all}\:\mathrm{non}- \\ $$$$\mathrm{negative}\:\mathrm{integers}\:\mathrm{subject}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{conditions}\:{n}_{\mathrm{0}} +{n}_{\mathrm{1}} +{n}_{\mathrm{2}} +{n}_{\mathrm{3}} +…+{n}_{{k}} ={n} \\ $$$$\mathrm{and}\:{n}_{\mathrm{1}} +\mathrm{2}{n}_{\mathrm{2}} +\mathrm{3}{n}_{\mathrm{3}} +…+{kn}_{{k}} ={p}. \\ $$

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