nhị thức niu tơn (3)

Dat $A=C^1_{2n+1}+C^2_{2n+1}+....+C^n_{2n+1}=2^{20}-1$

Ta co:$C^0_{2n+1}=C^{2n+1}_{2n+1}=1; C^1_{2n+1}=C^{2n}_{2n+1};C^2_{2n+1}=C^{2n-1}_{2n+1};....;C^n_{2n+1}=C^{n-1}_{2n+1}$

Do do :$2A+2=C^0_{2n+1}+C^1_{2n+1}+C^2_{2n+1}+...+C^n_{2n+1}+C^{n+1}_{2n+1}+...+C^{2n+1}_{2n+1}$

                              $=(1+1)^{2n+1}$

$\Rightarrow  A=\frac{2^{2n+1}-2}{2}=2^{2n}-1=2^{20}-1\Leftrightarrow 2n=20\Leftrightarrow n=10$

Ta co khai trien:$(\frac{1}{x^4}+x^7)^{10}=\sum^{10}_{k=0} C^k_{10}(\frac{1}{x^4})^{10-k}(x^7)^k=\sum^{10}_{k=0} C^{k}_{10}x^{11k-40}$

$x^{26}$ tuong ung voi $11k-40=26\Leftrightarrow k=6$

Vay he so cua $x^{26}$ la $C^6_{10}=210$