六年级奥数题1/1*2+1/2*3 +...+1/n(n+1)>1921/2001
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六年级奥数题1/1*2+1/2*3 +...+1/n(n+1)>1921/2001
结果大于2001分之1921,
结果大于2001分之1921,
![六年级奥数题1/1*2+1/2*3 +...+1/n(n+1)>1921/2001](/uploads/image/z/8664714-18-4.jpg?t=%E5%85%AD%E5%B9%B4%E7%BA%A7%E5%A5%A5%E6%95%B0%E9%A2%981%2F1%2A2%2B1%2F2%2A3+%2B...%2B1%2Fn%28n%2B1%29%3E1921%2F2001)
原式=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.+1/n-1/(n+1)
=1-1/(n+1)=n/(n+1)〉1921/2001,用1同时减去两边
推出:1/(n+1)
=1-1/(n+1)=n/(n+1)〉1921/2001,用1同时减去两边
推出:1/(n+1)
证明不等式:(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n
六年级奥数题1/1*2+1/2*3 +...+1/n(n+1)>1921/2001
2^n/n*(n+1)
[3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简
化简(n+1)(n+2)(n+3)
计算:n(n+1)(n+2)(n+3)+1
lim[(n+3)/(n+1))]^(n-2) 【n无穷大】
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^2 n 1 ) 2/(n^2 n 2) 3/(n^2 n 3) ……n/(n^2 n n)) 当N越于无穷大
1 + (n + 1) + n*(n + 1) + n*n + (n + 1) + 1 = 2n^2 + 3n + 3
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