Sam Tiểu Thư bn đăng câu hỏi sai nơi rồi nha

\(\frac{a^5}{5}+\frac{a^3}{3}+\frac{7a}{15}\left(n\Rightarrow a\text{ }nha\right)=\frac{a^5}{5}+\frac{a^3}{3}+\frac{7a}{15}=\frac{a^5}{5}+\frac{a^3}{3}+\frac{15a-5a-3a}{15}=\frac{a^5-a}{5}+\frac{a^3-a}{3}+\frac{15a}{15}=\frac{a^5-a}{5}+\frac{a^3-a}{3}+a;a^k-a⋮k\left(a\in Z;1< k\in N\right)\left(fecmat\right)\Rightarrow\left\{{}\begin{matrix}a^5-a⋮5\\a^3-a⋮3\end{matrix}\right.\Rightarrow dpcm\)

\(\frac{a}{12}+\frac{a^2}{8}+\frac{a^3}{24}\left(n\Rightarrow a\text{ nha}\right)=\frac{a^3+3a^2+2a}{24}=\frac{\left(a+2\right)\left(a+1\right)a}{24}.a=2k\left(k\in N\right)\Rightarrow;\frac{a\left(a+1\right)\left(a+2\right)}{24}=\frac{2k.\left(2k+1\right)\left(2k+2\right)}{24}=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\Leftrightarrow k\left(k+1\right)\left(2k+1\right)⋮6\)

đáp án là đpcm

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

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

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

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

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

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

Lập luận là ra

Theo t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{ab+ac}{2}=\dfrac{bc+ba}{3}=\dfrac{ca+cb}{4}\)

\(=\dfrac{ab+ac+bc+ba-ca-cb}{2+3-4}=\dfrac{2ab}{1}\) \(\left(1\right)\)

\(=\dfrac{bc+cb+bc+ba-ab-ac}{3+4-2}=\dfrac{2bc}{5}\left(2\right)\)

\(=\dfrac{ab+ac+ca+cb-bc-ba}{2+4-3}=\dfrac{2ac}{3}\)\(\left(3\right)\)

Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow\dfrac{2ab}{1}=\dfrac{2bc}{5}=\dfrac{2ac}{3}\)

\(\dfrac{2ab}{1}=\dfrac{2bc}{5}\Leftrightarrow\dfrac{a}{1}=\dfrac{c}{15}\) \(\Leftrightarrow\dfrac{a}{3}=\dfrac{c}{15}\left(I\right)\)

\(\dfrac{2bc}{5}=\dfrac{2ac}{3}\Leftrightarrow\dfrac{b}{5}=\dfrac{a}{3}\left(II\right)\)

Từ \(\left(I\right)+\left(II\right)\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{15}\left(đpcm\right)\)