`A=n/3+n^2/2+n^3/6`

`=(n^3+3n^2+2n)/6`

`=(n(n^2+3n+2))/6`

`=(n(n+1)(n+2))/6`

Vì `n(n+1)(n+2)` là tích 3 số nguyên liên tiếp

`=>n(n+1)(n+2) vdots 6`

`=>(n(n+1)(n+2))/6 in Z(forall x in Z)`

đáp án là đpcm

a) Ta có \(A=\dfrac{n-5}{n-3}=\dfrac{n-3-2}{n-3}=1-\dfrac{2}{n-3}\). Để \(A\inℤ\) thì \(\dfrac{2}{n-3}\inℤ\) hay \(n-3\) là ước của 2. Suy ra \(n-3\in\left\{\pm1;\pm2\right\}\). 

Nếu \(n-3=1\Rightarrow n=4\); \(n-3=-1\Rightarrow n=2\); \(n-3=2\Rightarrow n=5\); \(n-3=-2\Rightarrow n=1\). Vậy để \(A\inℤ\) thì \(n\in\left\{1;2;4;5\right\}\)

 \(A=\dfrac{n+4}{n+1}\) làm tương tự.

b) Dễ thấy các số ở mẫu có thể viết dưới dạng:

\(10=1+2+3+4=\dfrac{4\left(4+1\right)}{2}=\dfrac{4.5}{2}\)

\(15=1+2+3+4+5=\dfrac{5\left(5+1\right)}{2}=\dfrac{5.6}{2}\)

\(21=1+2+...+6=\dfrac{6\left(6+1\right)}{2}=\dfrac{6.7}{2}\)

...

\(120=1+2+...+15=\dfrac{15\left(15+1\right)}{2}=\dfrac{15.16}{2}\)

Do đó \(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\) 

\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(A=2\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{16-15}{15.16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(A=\dfrac{3}{8}\)

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

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

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

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

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

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

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

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

Lập luận là ra

1) \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(3n+5-\dfrac{3}{n}\right)}{-n\left(1-\dfrac{5}{n}\right)}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n+5-\dfrac{3}{n}}{-\left(1-\dfrac{5}{n}\right)}\)

\(=\left[{}\begin{matrix}-\infty\left(n\rightarrow+\infty\right)\\+\infty\left(n\rightarrow-\infty\right)\end{matrix}\right.\)

Bài 2,3 tương tự, bạn tự làm nhé!

a, (n+3)²-(n-1)²

= n²+6n+9-n²+2n-1

= 8n + 8

= 8(n+1) chia hết cho 8

Ta có: \(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+...+\dfrac{11}{5^{12}}\)

\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+...+\dfrac{11}{5^{11}}\)

\(\Rightarrow5A-A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)

\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)

\(\Rightarrow20A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)

\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)

\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)

\(\Rightarrow A< \dfrac{1}{16}\)