cong thuc nhi thuc newton

T

thien0526

ta có
[TEX]11^{10}=(10+1)^{11}=C_{10}^0+C_{10}^1.10+C_{10}^2.10^2+C_{10}^3.10^3+...+C_{10}^8.10^8+C_{10}^9.10^9+C_{10}^{10}.10^{10}=1+100+C_{10}^2.100+C_{10}^3.10.100+...+C_{10}^9.10^7.100+C_{10}^{10}.10^8.100[/TEX]
[TEX]\Rightarrow 11^{10}-1=100+C_{10}^2.100+C_{10}^3.10.100+...+C_{10}^9.10^7.100+C_{10}^{10}.10^8.100=100.(C_{10}^2+C_{10}^3.10+...+C_{10}^9.10^7+C_{10}^{10}.10^8)\vdots\ 100[/TEX]
(Vì [TEX]C_{10}^2;C_{10}^3.10;...;C_{10}^9.10^7;C_{10}^{10}.10^8[/TEX]đều là số nguyên dương)
Câu b bạn biến đổi tương tự nha

[TEX]c) (1+\sqrt{10})^{100}-(1-\sqrt{10})^{100}[/TEX]
[TEX]= \sum_{k=0}^{100} C_{100}^k\sqrt{10}^{k}-\sum_{k=0}^{100} C_{100}^k(-1)^k.\sqrt{10}^k[/TEX]
[TEX]=\sum_{k=0}^{49} C_{100}^{2k+1}.2.\sqrt{10}^{2k+1}[/TEX]
[TEX]= \sum_{k=0}^{49} C_{100}^{2k+1}.2.\sqrt{10}^{2k}.\sqrt{10}[/TEX]
[TEX]\Rightarrow \sqrt{10}[ (1+\sqrt{10})^{100}-(1-\sqrt{10})^{100}][/TEX]
[TEX]= \sqrt{10}.(\sum_{k=0}^{49} C_{100}^{2k+1}.2.\sqrt{10}^{2k}.\sqrt{10})[/TEX]
[TEX]= 10.(\sum_{k=0}^{49} C_{100}^{2k+1}.2.10^k[/TEX] là một số nguyên
 
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