Bài 104631

$(x+2)^n=\sum_{k=0}^nC^k_nx^{n-k}2^k=C^0_nx^n+2C^1_nx^{n-1}+2^2C^2_nx^{n-2}+2^3C^3_nx^{n-3}+…+2^nC^n_n$

Thay $x=1$ vào hai vế:

$3^n=C^0_n+2C^1_n+2^2C^2_n+2^3C^3_n+…+2^nC^n_n=243 \Leftrightarrow 3^n=3^5 \Leftrightarrow n=5$

b.Xét khai triển:

$(x+1)^{2n}=\sum_{k-0}^{2n}C^k_{2n}x^k=C^0_{2n}+C^1_{2n}x+C^2_{2n}x^2+C^3_{2n}x^3+…+C^{2n-1}_{2n}x^{2n-1}+C^{2n}_{2n}x^{2n}$

Thay $x=1$ vào 2 vế:

$2^{2n}=C^0_{2n}+C^1_{2n}+C^2_{2n}+C^3_{2n}+…+C^{2n-1}_{2n}+C^{2n}_{2n}$ (1)

Thay $x=-1$ vào 2 vế:

$0=C^0_{2n}-C^1_{2n}+C^2_{2n}-C^3_{2n}+…-C^{2n-1}_{2n}+C^{2n}_{2n}$ (2)

Lấy (1) – (2) với vế theo vế:

$2^{2n}=2(C^1_{2n}+C^3_{2n}+C^5{2n}+…+C^{2n-1}_{2n}) \Leftrightarrow 2^{2n}=2.2048$

$\Leftrightarrow 2^{2n}=4096=2^{12} \Leftrightarrow 2n=12 \Leftrightarrow n=6$