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?
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?
2^n/n*(n+1)
已知等差数列{an}的前n项和为Sn,且(2n-1)Sn+1 -(2n+1)Sn=4n²-1(n∈N*)
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
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
证明:1+2C(n,1)+4C(n,2)+...+2^nC(n,n)=3^n .(n∈N+)
如果正整数n使得[n/2]+[n/3]+[n/4]+[n/5]+[n/6]=69,则n=
n(n+1)(n+2)等于多少?
(2^n+4^n+6^n+8^n)^(1/n)当n趋于无穷时的极限
如果正整数n使得[n/2]+[n/3]+[n/4]+[n/5]+[n/6]=69,则n为( ).([ n ]表示不超过n
[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简