数学式测试：
( 1 + x n + 1 ) = ( 1 − x ) ( 1 + x + x 2 + x 3 + . . . + x n ) {\displaystyle (1+x^{n+1})=(1-x)(1+x+x^{2}+x^{3}+...+x^{n})} So 1 + x n + 1 1 − x = 1 + x + x 2 + x 3 + . . . + x n {\displaystyle {\frac {1+x^{n+1}}{1-x}}=1+x+x^{2}+x^{3}+...+x^{n}} generates
1, 1, 1, ..., 1, 0, 0, 0, ...
Extending (a), 1 = ( 1 − x ) ( 1 + x + x 2 + x 3 + . . . ) {\displaystyle 1=(1-x)(1+x+x^{2}+x^{3}+...)} So 1 1 − x = 1 + x 2 + x 3 + . . . {\displaystyle {\frac {1}{1-x}}=1+x^{2}+x^{3}+...} generates
1, 1, 1, ...
So
generates
So
generates
