3^n-1 - 3^n + 3^n+1 = 567看起来蛮简单,但就是解不出来...
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 12:00:40
3^n-1 - 3^n + 3^n+1 = 567看起来蛮简单,但就是解不出来...
3^n-1 - 3^n + 3^n+1 = 567
看起来蛮简单,但就是解不出来...
3^n-1 - 3^n + 3^n+1 = 567看起来蛮简单,但就是解不出来...
3^(n-1) - 3^n + 3^(n+1)
=3^n/3-3^n+(3^n)*3
=(1/3-1+3)*3^n
=(7/3)3^n
=567
所以:3^n=567*(3/7)=3*81=3^5
所以:n=5
5
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
判断 当n>1时,n*n*n>3n.( )
[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简
[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简
化简n分之n-1+n分之n-2+n分之n-3+.+n分之1
化简n分之n-1+n分之n-2+n分之n-3+.+n分之1
化简(n+1)(n+2)(n+3)
当n为正偶数,求证n/(n-1)+n(n-2)/(n-1)(n-3)+...+n(n-2).2/(n-1)(n-3)...1=n
lim2^n +3^n/2^n+1+3^n+1
lim(n+3)(4-n)/(n-1)(3-2n)
lim(n^3+n)/(n^4-3n^2+1)
3(n-1)(n+3)-2(n-5)(n-2)
n(n+1)(n+2)(n+3)+1 因式分解
n(n+1)(n+2)(n+3)+1等于多少
lim[(n+3)/(n+1))]^(n-2) 【n无穷大】
lim(2^n+3^n)^1
(n趋向无穷)
级数n/(n+1)(n+2)(n+3)和是多少
n*1+n*2+n*3+n*4.求公式