eln(2)+22⋅2+∑n=0∞(21)n+∫012dx+tr([1001])+∣{0}∪{1}∣+∑k=01P(X=k)
平方差公式推导
a2−b2
=a[b+(a−b)]−b2
=ab+a(a−b)−b2
=b(a−b)+a(a−b)
=(a+b)(a−b)
四面体数公式推导
令n=0,1,2,3,4,...
an=0,1,4,10,20,...
设数列b为数列a相邻两个元素之差则:
bn=1,3,6,10,...
再设数列c为数列b相邻两个元素之差则:
cn=2,3,4,...
而数列c是一个等差数列
cn=n+2
由递推关系可得
c0+c1+...+cn=bn−1−b0
所以bn=21(n+1)(n+2)
而b0+b1+...+bn=an+1−a0
an+1=1+3+...+21n2+23n+1
=21(0+1+4+...+n2)+23(0+1+2+...+n)+(n+1)
=121n(n+1)(2n+1)+43(n+1)+(n+1)
=121n(n+1)(2n+1)+47(n+1)
化简得:61n(n+1)(n+2)