x^n-y^n
當(dāng)n=2時,x^2-y^2=(x+y)(x-y) ,所以 (x^2-y^2)/(x+y)=x-y=f ,f為整數(shù).
設(shè)當(dāng)n=2k 之前都成立 即 x^2k-y^2k 能被x+y整除,即有 (x^2k-y^2k)/(x+y)=g ,g 為整數(shù)
當(dāng)n=2(k+1) 時,
x^2(k+1)-y^2(k+1)
=x^2* x^2k - y^2* y^2k
所以
[ x^2(k+1)-y^2(k+1) ] / (x+y)
=(x^2* x^2k) / (x+y)- (y^2* y^2k)/ (x+y) 說明:將(x+y)移進去,
=(x^2/(x+y)* x^2k) - (y^2* y^2k)/ (x+y) 說明:由 (x^2-y^2)/(x+y)=f 知道 x^2/(x+y)=f-y^2/(x+y)
= [ f-y^2/(x+y) ]* x^2k - (y^2* y^2k)/ (x+y)
=f* x^2k - y^2/(x+y)* x^2k - (y^2* y^2k)/ (x+y)
=f* x^2k - y^2* [ x^2k/(x+y) - y^2k/ (x+y)] 說明:n=2k 時 我們已經(jīng)有(x^2k-y^2k)/(x+y)=g
=f* x^2k - y^2*g
f是整數(shù),g是整數(shù),x^2k,y^2都是整數(shù),
所以[ x^2(k+1)-y^2(k+1) ] / (x+y) = h ,h為整數(shù).
即n=2(k+1) 時也成立,所以對所有n=2k (k=1,2,3...) 都成立.
考慮到太亂可能看不懂,我加了說明.
另外,事實上：有
x^n-y^n=(x+y)[x^(n-1)-x^(n-2)y+x^(n-3)y^2-…………+xy^(n-2)-y^(n-1)]
不過跟題目無關(guān),因為題目要歸納法證明.