已知 1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6,则1^2+3^2+...+（2n+1）^2=

来源：学生作业帮编辑：搜狗做题网作业帮分类：数学作业时间：2024/06/25 17:26:09

已知 1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6,则1^2+3^2+...+（2n+1）^2=?

2^2+4^2+...+(2n)^2
=2^2(1^2+2^2+..+n^2)
= 4n(n+1)(2n+1)/6
1^2+2^2+...+(2n+1)^2 = (2n+1)(2n+2)(2n+3)/6
1^2+3^2+...+（2n+1）^2
= 1^2+2^2+...+(2n+1)^2 - ( 2^2+4^2+...+(2n)^2 )
=(2n+1)(2n+2)(2n+3)/6 - 4n(n+1)(2n+1)/6
=[(n+1)(2n+1)/3 ] ( (2n+3) - 2n )
= (n+1)(2n+1)

证明不等式：(1/n)^n+(2/n)^n+(3/n)^n+.+(n/n)^n [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 证明(1+2/n)^n>5-2/n（n属于N+,n>=3) 化简(n+1)(n+2)(n+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? 1\n(n+3)+1\(n+3)(n+6)+1\(n+6)(n+9)=1\2 n+18 n为正整数,求n的值证明：1+2C(n,1)+4C(n,2)+...+2^nC(n,n)=3^n .(n∈N+) 已知递推公式f(n)=(n-1)(n-2)[f(n-2)+f(n-3)+(n-3)*f(n-4)] (n>4)求通项公式阶乘(2n-1)!=（2n)!/(2^n*n! 计算：n(n+1)(n+2)(n+3)+1