裁决者图哈特 wow:用累加法求数列1 1 2 4 7.......的第n项

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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+……+an=n(a1+an)/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:    an- an-1 = n-2............................n-1                     第(n-1)项减去第(n-2)项的差为(n-3)               an-1-an-2 = n-3 .........................n-2                                                                           .........                                                              ..........                      第 5 项减去第 4 项的差等于 3                            a- a4    =   3    ...........................4                                       第 4 项减去第 3 项的差等于 2                             a- a3    =   2   ............................3                                      第 3 项减去第 2 项的差等于 1                             a-  a  =   1   ............................2                      第 2 项减去第 1 项的差等于 0                             a-  a1   =   0   ............................1                  将所得结果(共有n-1个式子)左右两边相加得:                                                                                                       an- a1   =  0+1+2+3+4+……+n-3 +n-2 
                                =(n-1)[0+(n-2)]/2
                                                                =  (n-1)(n-2)/2                          所以    an=a +   (n-1)(n-2)/2 = 1+ (n-1)(n-2)/2