Giải thích các bước giải:
Với $n=1\to u_1=7\quad\vdots\quad 7$
Giải sử $n=k\to u_k=k^7+6k \quad\vdots\quad 7$
Ta cần chứng minh $u_{k+1}\quad\vdots\quad 7$
Thật vậy ta có:
$u_{k+1}=(k+1)^7+6(k+1)$
$\to u_{k+1}-u_k=(k+1)^7+6(k+1)-(k^7+6k)$
$\to u_{k+1}-u_k=(k+1)^7-k^7+6$
$\to u_{k+1}-u_k=k^7+7k^6+21k^5+35k^4+35k^3+21k^2+7k+1-k^7+6$
$\to u_{k+1}-u_k=7k^6+21k^5+35k^4+35k^3+21k^2+7k+7$
$\to u_{k+1}-u_k=7(k^6+3k^5+5k^4+5k^3+3k^2+k+1)$
$\to u_{k+1}-u_k\quad\vdots\quad 7$
Mà $u_k\quad\vdots\quad 7$
$\to u_{k+1}\quad\vdots\quad 7$
$\to u_n=n^7+6n\quad\vdots\quad 7$
