问题描述：
如何运用 倒序相加法 证明二项式定理各项系数和为2的n次方
问题解答：
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∵(a+b)^n=∑(k=0,n)ℂnk‧a^(n−k) b^k
2^n=(1+1)^n
=∑(k=0,n)ℂnk‧1^(n−k) 1^k
=∑(k=0,n)ℂnk
=ℂn0+ℂn1+ℂn2+…+ℂnk−2+ℂnk−1+ℂnk
S1=ℂn0+ℂn1+ℂn2+…+ℂnn−2+ℂnn−1+ℂnn
S2=ℂnn+ℂnn−1+ℂnn−2+…+ℂn2+ℂn1+ℂn0
S1+S2=[ℂn0+ℂnn]+[ℂn1+ℂnn−1)]…+[ℂnn−1+ℂn1]+[ℂnn+ℂn0]
=2[ℂn0+ℂn1+ℂn2+…+ℂnn−2+ℂnn−1+ℂnn]=2S1 (or S2 )
=2[ℂn0‧1^n‧1^0+ℂn1‧1^(n−1)‧1^1+ℂn2‧1^(n−2)‧1^2+…+ℂn(n−1)‧1^1‧1^(n−1)+ℂnn‧1^0‧1^n ]
=2‧(1+1)^n=2‧2^n
∴S1=2^n
2^n=(1+1)^n
=∑(k=0,n)ℂnk‧1^(n−k) 1^k
=∑(k=0,n)ℂnk
=ℂn0+ℂn1+ℂn2+…+ℂnk−2+ℂnk−1+ℂnk
S1=ℂn0+ℂn1+ℂn2+…+ℂnn−2+ℂnn−1+ℂnn
S2=ℂnn+ℂnn−1+ℂnn−2+…+ℂn2+ℂn1+ℂn0
S1+S2=[ℂn0+ℂnn]+[ℂn1+ℂnn−1)]…+[ℂnn−1+ℂn1]+[ℂnn+ℂn0]
=2[ℂn0+ℂn1+ℂn2+…+ℂnn−2+ℂnn−1+ℂnn]=2S1 (or S2 )
=2[ℂn0‧1^n‧1^0+ℂn1‧1^(n−1)‧1^1+ℂn2‧1^(n−2)‧1^2+…+ℂn(n−1)‧1^1‧1^(n−1)+ℂnn‧1^0‧1^n ]
=2‧(1+1)^n=2‧2^n
∴S1=2^n
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