CMR với V \(n\ge1\) ta có:

5^2n-1.2^2n-1.5ⁿ⁺¹+3ⁿ⁺¹.2^2n-1 chia hết cho 38

cmr:

\(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{2n-1}{2n}\le\dfrac{1}{\sqrt{3n+1}}\left(\forall n\ge1\right)\)

CMR với mọi n > hoặc băng 1 ta có

5^2n-1.2^2n-15ⁿ⁺¹+3ⁿ⁺¹.2^2n-1 chia hết cho 38

Chứng minh rằng \(2^{n+1}.5^{2n-1}+2^{2n-1}.3^{n+1}⋮38\left(n\in N,n\ge1\right)\)

CMR \(\forall n\in\)N* ta có

\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+...+\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\)

Cmr: \(5^{2n-1}.2^{n+1}+2^{2n-1}.3^{n+1}⋮38\) ( n ∈ N* )

Chứng minh rằng: \(5^{2n-1}.2^{2n-1}.5^{n+1}+3^{n+1}.2^{2n-1}=2^n\left(5^{2n-1}.10+9.6^{n-1}\right)\)

Với \(n\ge1\)

Chứng minh rằng \(5^{n-1}.2^{2n-1}.5^{n+1}+3^{n+1}.2^{2n-1}=2^n\left(5^{2n-1}.10+9.6^{n-1}\right)\)

Với \(n\ge1\)

CMR voi moi so tu nhien n thi

A=5^2n+1*2^n+2+3^n+2*2^2n+1 chia het cho 38