a, Ta có: \(\frac{n^5}{5}+\frac{n^3}{3}+\frac{7n}{15}=\frac{n^5-n}{5}+\frac{n}{5}+\frac{n^3-n}{3}+\frac{n}{3}+\frac{7n}{15}\) 

\(=\frac{n^5-n}{5}+\frac{n^3-n}{3}+n\) 

Chứng minh \(n^5-n⋮5\Rightarrow\frac{n^5-n}{5}\in Z\) 

                   \(n^3-n⋮3\Rightarrow\frac{n^3-n}{3}\in Z\)

\(\Rightarrow\frac{n^5-n}{5}+\frac{n^3-n}{3}+n\in Z\) 

=> Đpcm 

b, Tương tự dùng tính chất chia hết

\(B=\frac{n^4}{24}+\frac{n^3}{4}+\frac{11n^2}{24}+\frac{n}{4}\)

\(B=\frac{n^4+6n^3+11n^2+6n}{24}\)

\(B=\frac{n^4+2n^3+4n^3+8n^2+3n^2+6n}{24}\)

\(B=\frac{n^3\left(n+2\right)+4n^2\left(n+2\right)+3n\left(n+2\right)}{24}\)

\(B=\frac{\left(n^3+n^2+3n^2+3n\right)\left(n+2\right)}{24}\)

\(B=\frac{n\left(n+1\right)\left(n+3\right)\left(n+2\right)}{24}\)

Lập luận là ra

Sam Tiểu Thư bn đăng câu hỏi sai nơi rồi nha

môn anh mà bạn đâu phải môn toán

đáp án là đpcm

\(S=\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{1}{16}\right)+...+\left(1-\dfrac{1}{n^2}\right)\\ S=\left(1+1+...+1\right)-\left(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\right)\\ S=n-1-\left(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}\right)< n-1\)

Lại có \(\dfrac{1}{4}+\dfrac{1}{9}+..+\dfrac{1}{n^2}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\)

\(\Rightarrow S>n-1-1=n-2\\ \Rightarrow n-2< S< n-1\\ \Rightarrow S\notin N\)