就是将一项写成差的形式,然后相互抵消即可
1/(3n-1)(3n+2)=(1/3)[1/(3n-1)-1/(3n+2)]
Sn=(1/3)[1/2-1/5+1/5-1/8+.+1/(3n-2)-1/(3n+2)]
=(1/3)[1/2-1/(3n+2)]
=(1/3)[(3n)/(6n+4)]
=n/(6n+4)裂项求和公式是什么？为什么可以=(1/3)[1/(3n-1)-1/(3n+2)]没有公式啊。 具体题具体分析。如果是等差数列{an}的相邻两项 1/an*a(n+1)=(1/d)[1/an-1/a(n+1)]