a+1b+1 ∑i=0ni2=(2n+1)(n+1)n6 ∏i=1ni=n! ∫ni=0 (n+m+1m) limx→0f(x+Δx)−f(x)Δx sin(α+β)=sinαcosβ+cosαsinβ QYQ2JrZXG+CzD−1