Giải thích các bước giải:
a.Ta có: $9^{2n+1}+1=9^{2n+1}+1^{2n+1}\quad\vdots\quad 9+1=10$ vì $2n+1$ lẻ
b.Ta có:
$3^{4n+1}+2=3^{4n}\cdot 3+2$
$\to 3^{4n+1}+2=3^{4n}\cdot 3-3+5$
$\to 3^{4n+1}+2=81^{n}\cdot 3-3+5$
$\to 3^{4n+1}+2=(81^{n}-1)\cdot 3+5$
Vì $81^n-1\quad\vdots\quad 81-1\quad\vdots\quad 5$
$\to (81^{n}-1)\cdot 3+5\quad\vdots\quad 5$
$\to 3^{4n+1}+2\quad\vdots\quad 5$
c.Ta có: $S_n=4^{n+1}+5^{2n-1}$
Nếu $n=1\to S_1=4^{1+1}+5^{2\cdot 1-1}=21\quad\vdots\quad 21$
Giả sử $n=k\to S_k=4^{k+1}+5^{2k-1}\quad\vdots\quad 21, k\in N^*, k\ge 2$
Ta cần chứng minh $n=k+1\to S_{k+1}\quad\vdots\quad 21$
Thật vậy ta có:
$S_{k+1}=4^{(k+1)+1}+5^{2(k+1)-1}$
$\to S_{k+1}=4\cdot 4^{k+1}+5^{2k+1}$
$\to S_{k+1}=4\cdot 4^{k+1}+5^{2k-1+2}$
$\to S_{k+1}=4\cdot 4^{k+1}+5^{2k-1}\cdot 25$
$\to S_{k+1}=4\cdot 4^{k+1}+5^{2k-1}\cdot 4+5^{2k-1}\cdot 21$
$\to S_{k+1}=4\cdot (4^{k+1}+5^{2k-1})+5^{2k-1}\cdot 21$
Vì $4^{k+1}+5^{2k-1}\quad\vdots\quad 21, 21\quad\vdots\quad 21$
$\to 4\cdot (4^{k+1}+5^{2k-1})+5^{2k-1}\cdot 21\quad\vdots\quad 21$
$\to S_{k+1}\quad\vdots\quad 21$
$\to đpcm$
Vậy $4^{n+1}+5^{2n-1}\quad\vdots\quad 21,\quad\forall n\in N^*$
