裂项相消法 隔项相消原式=1/2[(1-1/3)+(1/2-1/4)+(1/3-1/5)+.+(1/n-1/(n+2))

裂项相消法常见公式1/n(n+1)(n+2)=? 裂项相消法求和（1）an=1/(2n+1)(2n+3)(2)an=5/n(n+2)(3)an=1/(n+1)(n+2)( 数列求和中拆项相消法怎么拆,有什么公式吗,比如：1/(2n-1)(2n+1) 证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n 求数列1/1+根号2 .1/根号2+根号3.1/根号n+根号n+1 的前几项和Sn,用裂项相消法做 证明(1+2/n)^n>5-2/n（n属于N+,n>=3) [3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简 2^n/n*(n+1) 1 + (n + 1) + n*(n + 1) + n*n + (n + 1) + 1 = 2n^2 + 3n + 3 Sn=n(n+2)(n+4)的分项等于1/6[n(n+2)(n+4)(n+5)-(n-1)n(n+2)(n+4)]吗? 化简：1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4) (n+1)(n+2)/1 +(n+2)(n+3)/1 +(n+3)(n+4)/1