数列\(\left\{a_n\right\}\)，\(\left\{b_n\right\}\)、\(\left\{c_n\right\}\)满足：\[\begin{split}b_n&=a_n-a_{n+2},\\c_n&=a_n+2a_{n+1}+3a_{n+2}.\end{split}\]若数列\(\left\{c_n\right\}\)为等差数列，且\(b_n\leqslant b_{n+1}\)（\(n=1,2,3,\cdots\)），求证：数列\(\left\{a_n\right\}\)是等差数列．
根据题意有\[c_n-c_{n+2}=b_n+2b_{n+1}+3b_{n+2},\]又数列\(\left\{c_n\right\}\)为等差数列，因此\[c_n-c_{n+2}=c_{n+1}-c_{n+3},\]于是\[b_n+2b_{n+1}+3b_{n+2}=b_{n+1}+2b_{n+2}+3b_{n+3},\]即\[\left(b_n-b_{n+1}\right)+2\left(b_{n+1}-b_{n+2}\right)+3\left(b_{n+2}-b_{n+3}\right)=0,\]结合\(b_n\leqslant b_{n+1}\)（\(n=1,2,3,\cdots\)）可得\(\left\{b_n\right\}\)为常数列．
不妨设\(b_n=-2d\)，于是数列\(\left\{a_n\right\}\)的奇数项和偶数项分别构成以\(2d\)为公差的等差数列，因此只需要证明\(a_2-a_1=d\)即可．
根据题意，有\[\begin{split}c_1&=a_1+2a_2+3a_3=4a_1+2a_2+6d,\\c_2&=a_2+2a_3+3a_4=2a_1+4a_2+10d,\\c_3&=a_3+2a_4+3a_5=4a_1+2a_2+18d,\end{split}\]而\[2c_2=c_1+c_3,\]于是不难得到\[a_2-a_1=d,\]
因此数列\(\left\{a_n\right\}\)是等差数列．