用累加法求数列1 1 2 4 7.......的第n项

例：已知数列1  1  2  4  7  11  16......

求：第10项是什么数，第n项是什么？（等差数列求和公式：[1+2+3+4+5+6+……+a_n=n(a₁+a_n)/2  ] ）

解：观察数列1  1  2  4  7  11........可知，  将数列的后一项减去前一项就得到一列新的数列为：0  1  2  3  4  5  6.......

       所以有  第     n   项减去第(n-1)项的差为（n-2）       即:    a_n- a_n-1 = n-2............................n-1

                     第（n-1)项减去第(n-2)项的差为（n-3）               a_n-1-a_n-2 = n-3 .........................n-2                                 

                                          .........                                                              ..........

                      第 5 项减去第 4 项的差等于 3                            a₅  - a₄    =   3    ...........................4                 

                      第 4 项减去第 3 项的差等于 2                             a₄  - a₃    =   2   ............................3               

                      第 3 项减去第 2 项的差等于 1                             a₃  -  a₂   =   1   ............................2

                      第 2 项减去第 1 项的差等于 0                             a₂  -  a₁   =   0   ............................1

                  将所得结果（共有n-1个式子）左右两边相加得：                                                    

                                                   a_n- a₁   =  0+1+2+3+4+……+n-3 +n-2 
                                =（n-1）[0+(n-2)]/2

                                                                =  (n-1)(n-2)/2

                          所以    a_n=a₁  +   (n-1)(n-2)/2 = 1+ (n-1)(n-2)/2