\(2^{n+1}:4^2=1024\)

\(2^{n+1}:16=1024\)

\(2^{n+1}=1024.16\)

\(2^{n+1}=16384\)

\(2^{n+1}=2^{14}\)

\(\Rightarrow n+1=14\)

\(\Rightarrow n=13\)

vậy \(n=13\)

2ⁿ⁺¹   : 4² =1024.

=>2ⁿ⁺¹:(2²)²=2¹⁰

=>2ⁿ⁺¹:2⁴=2¹⁰

=>2^n+1-4=2¹⁰

=>n+1-4=10

=>n+1=14

=>n=13.

Đáp án là B

Ta có: \(3\cdot A_{n}^{n-2}+C_{n}^3=40\)

=>\(3\cdot\frac{n!}{\left(n-2\right)!}+\frac{n!}{\left(n-3\right)!\cdot3!}=40\)

=>\(3n\left(n-1\right)+\frac{n\left(n-1\right)\left(n-2\right)}{6}=40\)

=>\(18n\left(n-1\right)+n\left(n-1\right)\left(n-2\right)=240\)

=>\(n\left(n-1\right)\left(n+16\right)=240\)

=>n=4

=>2n=8

Số hạng tổng quát là: \(C_8^{k}\cdot\left(2x\right)^{8-k}\cdot\left(-\frac{1}{x}\right)^{k}=C_8^{k}\cdot2^{8-k}\cdot x^{8-k}\cdot\frac{\left(-1\right)^{k}}{x^{k}}\)

\(=C_8^{k}\cdot2^{8-k}\cdot\left(-1\right)^{k}\cdot x^{8-k\cdot2}\)

Hệ số chứa x^6 sẽ tương ứng với 8-2k=6

=>2k=2

=>k=1

=>Hệ số là \(C_8^1\cdot2^{8-1}\cdot\left(-1\right)^1=-8\cdot2^7=-2^{10}=-1024\)

=>Chọn A

giúp mk vi.trong đây ko có đáp án đ

sao lại C^3_2n+1 nhỉ

ta có \(\left(1+1\right)^{2n+1}=C_{2n+1}^0+C^1_{2n+1}+C^2_{2n+1}+...+C^{2n+1}_{2n+1}\)

\(-\left(1-1\right)^{2n+1}=-\left(C_{2n+1}^0-C^1_{2n+1}+C^2_{2n+1}-...-C^{2n+1}_{2n+1}\right)\)

\(\left(1+1\right)^{2n+1}-\left(1-1\right)^{2n+1}=C_{2n+1}^0+C^1_{2n+1}+C^2_{2n+1}+...+C^{2n+1}_{2n+1}-C_{2n+1}^0+C_{2n+1}^1-C_{2n+1}^2+....+C_{2n+1}^{2n+1}\)

\(2^{2n+1}=2C_{2n+1}^1+2C_{2n+1}^3+2C_{2n+1}^5+...+C_{2n+1}^{2n+1}=2.1024=2048\)

\(\Rightarrow n=5\)

\(\left(2-3x\right)^{10}\)

SHTQ \(C_{10}^k.2^{10-k}.\left(-3x\right)^k=C_{10}^k.2^{10-k}.-3^k.x^k\)

\(x^7\Rightarrow k=7\)

hệ số cần tìm \(C_{10}^7.2^3.\left(-3\right)^7=-2099520\)

2^n+1:4^2=1024

2^n+1       =1024.16

2^n            =16384-1

2^n            =16383

bài toán này đề sai nhé bn 

Ta có

\(\frac{4^{n+3}+17.2^{2n}}{9^{n+1}+7.3^{2n}}=\frac{2^{2n+6}+17.2^{2n}}{3^{2n+2}+7.3^{2n}}=\frac{2^{2n}.\left(2^6+17\right)}{3^{2n}.\left(3^2+7\right)}=\left(\frac{2}{3}\right)^{2n}.\frac{81}{16}=1\)

\(\Rightarrow\left(\frac{2}{3}\right)^{2n}.\frac{3^4}{2^4}=1\Rightarrow\left(\frac{2}{3}\right)^{2n}=\left(\frac{2}{3}\right)^4\Rightarrow2n=4\Rightarrow n=2\)