17The argument concludes as that of the base step.(n+1)2=(nω+εn+1)·(nω+εn+1)=n2+2nω·εn+1+1,so thatω·εn+1=1. By the base step,ω=εn+1, and the proof iscomplete.1.36(Star of David) Prove, for alln>r≥1, thatn−1r−1nr+1n+1r=n−1rnr−1n+1r+1.n−1r−1n−1rnr−1nrnr+1n+1rn+1r+1Solution.Using Pascal’s formula, one sees that both sides are equal to(n−1)!n!(n+1)!(r−1)!r!(r+1)!(n−r−1)!(n−r)!(n−r+1)!.1.37For all oddn≥1, prove that there is a polynomial
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Trigraph, Complex number, de Moivre, IIII, iiii UUU