\(\left(3^{n+1}-2.2^n\right)\left(3.3^n+2^{n+1}\right).3^{2n+2}+\left(8.2^{n-2}.3^{n+1}\right)^2\)

\(=\left(3^{n+1}-2^{n+1}\right)\left(3^{n+1}+2^{n+1}\right).3^{2n+2}+\left(2^{n+1}.3^{n+1}\right)^2\)

\(=\left(3^{2n+2}-2^{2n+2}\right).3^{2n+2}+2^{2n+2}.3^{2n+2}\)

\(=3^{2\left(2n+2\right)}-2^{2n+2}.3^{2n+2}+2^{2n+2}.3^{2n+2}\)

\(=3^{2\left(2n+2\right)}=\left(3^{2n+2}\right)^2\).

Ta thấy \(\left(3^{2n+2}\right)^2\)luôn là 1 số chính phương với mọi n\(\in\)N

Nên ta có ĐPCM.

Do 2 + 1 chia hết cho 3 nên theo bổ đề LTE ta có \(v_3\left(2^{3^n}+1\right)=v_3\left(2+1\right)+v_3\left(3^n\right)=n+1\).

Do đó \(2^{3^n}+1⋮3^{n+1}\) nhưng không chia hết cho \(3^{n+2}\).

\(\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{2}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}\)

\(=\frac{1}{2}\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]\)

Ta có đpcm. 

a^2 + b^2 + c^2= ab + bc + ca

2 ( a^2 + b^2 + c^2 ) = 2 ( ab + bc + ca)

2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca

a^2 + a^2 + b^2 + b^2 + c^2+ c^2 – 2ab – 2bc – 2ca = 0

a^2 + b^2 – 2ab + b^2 + c^2 – 2bc + c² + a² – 2ca = 0

(a^2 + b^2 – 2ab) + (b^2 + c^2 – 2bc) + (c^2 + a^2 – 2ca) = 0

(a – b)^2 + (b – c)^2 + (c – a)^2 = 0

Vì (a-b)^2 lớn hơn hoặc bằng 0 với mọi a và b 

     (b-c)^2  lớn hơn hoặc bằng 0 với mọi c và b

     (c-a)^2 lớn hơn hoặc bằng 0 với mọi a và c

=> (a-b)^2 =0  ; (b-c)^2=0 ; (c-a)^2=0

=> a=b ; b=c ; c=a

=>a=b=c

Lời giải:
Xét:

$M=1+10+....+10^n$

$10M=10+10^2+....+10^{n+1}$
$10M-M=10^{n+1}-1$

$M=\frac{10^{n+1}-1}{9}$

$A=M.(10^{n+1}+5)+1=\frac{(10^{n+1}-1)(10^{n+1}+5)}{9}+1$

$=\frac{10^{2n+2}+4.10^{n+1}-5+9}{9}$

$=\frac{10^{2n+2}+4.10^{n+1}+4}{9}$

$=\frac{(10^{n+1}+2)^2}{9}$

$=\left(\frac{10^{n+1}+2}{3}\right)^2$
Ta thấy: $10^{n+1}+2\equiv 1^{n+1}+2=3\equiv 0\pmod 3$

Do đó: $\frac{10^{n+1}+2}{3}\in\mathbb{N}$

Suy ra $A$ là scp.

Ta có: \(a_n=1+\frac{2^n\left[1.3.5...\left(2n-1\right)\right]}{\left(n+5\right)\left(n+6\right)...\left(2n\right)}\)

\(=1+\frac{2^n\left(2n\right)!}{\left[2.4.6..\left(2n\right)\right]\left[\left(n+5\right)\left(n+6\right)..\left(2n\right)\right]}\)

\(=1+\frac{\left(2n\right)!}{n!\left(n+5\right)\left(n+6\right)...\left(2n\right)}\)

\(=1+\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)\)

mặt khác \(1+\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)=\left(n^2+5n+5\right)^2\)

do đó a_n luôn là SCP

\(1-\dfrac{3}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)-3}{n\left(n+2\right)}=\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

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

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

\(=\dfrac{1}{n}.\dfrac{n+3}{4}=\dfrac{n+3}{4n}=\dfrac{1}{4}+\dfrac{3}{4n}>\dfrac{1}{4}\) (đpcm)