Tìm hệ số của $x^{10}$ trong khai triển: $$\left(1+x+x^2+x^3\right)^{15}$$
Khai triển nhị thức Newton: $(1+x+x^{2}+x^{3})^{15}$
= $C^{k}_{15}$. $(x+x^{2}+x^{3})^{k}$
= $C^{k}_{15}$. $C^{t}_{k}$. $x^{k-t}$ $(x^{2}+x^{3})^{t}$
= $C^{k}_{15}$. $C^{t}_{k}$. $x^{k-t}$. $C^{m}_{t}$ $(x^{})^{2m}$. $x^{3.(t-m)}$
 = $C^{k}_{15}$. $C^{t}_{k}$. $C^{m}_{t}$. $x^{k+2t-m}$
Khi đó ta cần tìm (k,t,m) sao cho: k+2t-m=0, k$\leqslant $ 15, t$\leqslant $k, m$\leqslant $t
=> (k,t,m)= $\left\{ {(4,3,0),(4,4,2),(5,3,1),(5,4,3),(5,5,5),(6,2,0),(6,3,2),(6,4,4),(7,2,1),(7,3,3),(8,1,0),(8,2,2),(9,1,1),(10,0,0)} \right\}$
Vậy hệ số của $x^{10}$ trong khai triển là: 1392456
chẳng hiểu gì luôn –  conheodat297 31-10-12 09:25 PM
chỉ dài hơn thôi chứ cách triển khai thì cơ bản mà! –  GreenmjlkTea FeelingTea 31-10-12 08:35 PM
bài này hóc búa quá –  giacmotrua297 31-10-12 08:31 PM

Theo khai triển nhị thức Newton ta có:

$(1+x+x^2+x^3)^{15}$

$=(1+x)^{15}(1+x^2)^{15}$

$=\left( \sum_{k=0}^{15}{C_{15}^kx^k}\right) \left( \sum_{k=0}^{15}{C_{15}^kx^{2k}}\right)$

Hệ số của $x^{10}$ trong khai triển trên là:

$C_{15}^0.C_{15}^{5}+C_{15}^2.C_{15}^4+C_{15}^4.C_{15}^3+C_{15}^6.C_{15}^2+C_{15}^8.C_{15}^1+C_{15}^{10}.C_{15}^0=1392456.$

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