\(1,\)

\(a,\) Sửa: \(A=10^n+72n-1⋮81\)

Với \(n=1\Leftrightarrow A=10+72-1=81⋮81\)

Giả sử \(n=k\Leftrightarrow A=10^k+72k-1⋮81\)

Với \(n=k+1\Leftrightarrow A=10^{k+1}+72\left(k+1\right)-1\)

\(A=10^k\cdot10+72k+72-1\\ A=10\left(10^k+72k-1\right)-648k+81\\ A=10\left(10^k+72k-1\right)-81\left(8k-1\right)\)

Ta có \(10^k+72k-1⋮81;81\left(8k-1\right)⋮81\)

Theo pp quy nạp 

\(\Rightarrow A⋮81\)

\(b,B=2002^n-138n-1⋮207\)

Với \(n=1\Leftrightarrow B=2002-138-1=1863⋮207\)

Giả sử \(n=k\Leftrightarrow B=2002^k-138k-1⋮207\)

Với \(n=k+1\Leftrightarrow B=2002^{k+1}-138\left(k+1\right)-1\)

\(B=2002\cdot2002^k-138k-138-1\\ B=2002\left(2002^k-138k-1\right)+276138k+1863\\ B=2002\left(2002^k-138k-1\right)+207\left(1334k+1\right)\)

Vì \(2002^k-138k-1⋮207;207\left(1334k+1\right)⋮207\)

Nên theo pp quy nạp \(B⋮207,\forall n\)

\(2,\)

\(a,\) Sửa đề: CMR: \(1\cdot2+2\cdot3+...+n\left(n+1\right)=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)

Đặt \(S_n=1\cdot2+2\cdot3+...+n\left(n+1\right)\)

Với \(n=1\Leftrightarrow S_1=1\cdot2=\dfrac{1\cdot2\cdot3}{3}=2\)

Giả sử \(n=k\Leftrightarrow S_k=1\cdot2+2\cdot3+...+k\left(k+1\right)=\dfrac{k\left(k+1\right)\left(k+2\right)}{3}\)

Với \(n=k+1\)

Cần cm \(S_{k+1}=1\cdot2+2\cdot3+...+k\left(k+1\right)+\left(k+1\right)\left(k+2\right)=\dfrac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}\)

Thật vậy, ta có:

\(\Leftrightarrow S_{k+1}=S_k+\left(k+1\right)\left(k+2\right)\\ \Leftrightarrow S_{k+1}=\dfrac{k\left(k+1\right)\left(k+2\right)}{3}+\left(k+1\right)\left(k+2\right)\\ \Leftrightarrow S_{k+1}=\dfrac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}\)

Theo pp quy nạp ta có đpcm

\(b,\) Với \(n=0\Leftrightarrow0^3=\left[\dfrac{0\left(0+1\right)}{2}\right]^2=0\)

Giả sử \(n=k\Leftrightarrow1^3+2^3+...+k^3=\left[\dfrac{k\left(k+1\right)}{2}\right]^2\)

Với \(n=k+1\)

Cần cm \(1^3+2^3+...+k^3+\left(k+1\right)^3=\left[\dfrac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\)

Thật vậy, ta có

\(1^3+2^3+...+k^3+\left(k+1\right)^3\\ =\left[\dfrac{k\left(k+1\right)}{2}\right]^2+\left(k+1\right)^3\\ =\dfrac{k^2\left(k+1\right)^2+4\left(k+1\right)^3}{4}=\dfrac{\left(k+1\right)^2\left(k^2+4k+4\right)}{4}\\ =\dfrac{\left(k+1\right)^2\left(k+2\right)^2}{4}=\left[\dfrac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\)

Theo pp quy nạp ta được đpcm

\(B=\left(1-\frac{3}{2.4}\right)\left(1-\frac{3}{3.5}\right)\left(1-\frac{3}{4.6}\right)...\left(1-\frac{3}{n\left(n+2\right)}\right)\)

\(=\frac{1.5}{2.4}.\frac{2.6}{3.5}.\frac{3.7}{4.6}...\frac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

\(=\frac{\left[1.2.3...\left(n-1\right)\right]\left[5.6.7...\left(n+3\right)\right]}{\left(2.3.4...n\right)\left[4.5.6...\left(n+2\right)\right]}\)

\(=\frac{n+3}{4n}< 2\left(đpcm\right)\)

Mashiro Shiina Akai Haruma GIÚP EM VỚI

3ⁿ⁺² - 2ⁿ⁺² +3ⁿ - 2ⁿ = 3ⁿ . 3² - 2ⁿ. 2² +3ⁿ -2ⁿ

                             = 3ⁿ(3²+1) - (2ⁿ.2² +2ⁿ)

                             =3ⁿ . 10 - 2ⁿ .5

                             =3ⁿ.10 - 2^n-1 .2 .5

                             = 3ⁿ.10 - 2^n-1 .10

                             = 10(3ⁿ - 2^n-1)

vì 10 chia hết cho 10 nên 10(3ⁿ-2^n-1) chia hết cho 10

                         =>  3ⁿ⁺² - 2ⁿ⁺² +3ⁿ -2ⁿ chia hết cho 10

Ai làm nhanh nhất mình sẽ **** xin cảm ơn các bạn mình đang cần gấp

2.

\(4n^3+n+3=4n^3+2n^2+2n-2n^2-n-1+4=2n\left(2n^2+n+1\right)-\left(2n^2+n+1\right)+4\)-Để \(\left(4n^3+n+3\right)⋮\left(2n^2+n+1\right)\) thì \(4⋮\left(2n^2+n+1\right)\)

\(\Leftrightarrow2n^2+n+1\in\left\{1;-1;2;-2;4;-4\right\}\) (do n là số nguyên)

*\(2n^2+n+1=1\Leftrightarrow n\left(2n+1\right)=0\Leftrightarrow n=0\) (loại) hay \(n=\dfrac{-1}{2}\) (loại)

*\(2n^2+n+1=-1\Leftrightarrow2n^2+n+2=0\) (phương trình vô nghiệm)

\(2n^2+n+1=2\Leftrightarrow2n^2+n-1=0\Leftrightarrow n^2+n+n^2-1=0\Leftrightarrow n\left(n+1\right)+\left(n+1\right)\left(n-1\right)=0\Leftrightarrow\left(n+1\right)\left(2n-1\right)=0\)

\(\Leftrightarrow n=-1\) (loại) hay \(n=\dfrac{1}{2}\) (loại)

\(2n^2+n+1=-2\Leftrightarrow2n^2+n+3=0\) (phương trình vô nghiệm)

\(2n^2+n+1=4\Leftrightarrow2n^2+n-3=0\Leftrightarrow2n^2-2n+3n-3=0\Leftrightarrow2n\left(n-1\right)+3\left(n-1\right)=0\Leftrightarrow\left(n-1\right)\left(2n+3\right)=0\)\(\Leftrightarrow n=1\left(nhận\right)\) hay \(n=\dfrac{-3}{2}\left(loại\right)\)

-Vậy \(n=1\)

1. \(x^2+y^2=z^2\)

\(\Rightarrow x^2+y^2-z^2=0\)

\(\Rightarrow\left(x-z\right)\left(x+z\right)+y^2=0\)

-TH1: y lẻ \(\Rightarrow x-z;x+z\) đều lẻ.

\(x+3z-y=x+z-y+2x\) chia hết cho 2. \(\Rightarrow\)Hợp số.

-TH2: y chẵn \(\Rightarrow\)1 trong hai biểu thức \(x-z;x+z\) chia hết cho 2.

*Xét \(\left(x-z\right)⋮2\):

\(x+3z-y=x-z+4z-y\) chia hết cho 2. \(\Rightarrow\)Hợp số.

*Xét \(\left(x+z\right)⋮2\):

\(x+3z-y=x+z+2z-y\) chia hết cho 2 \(\Rightarrow\)Hợp số.