Loading Documenta[n]=Πκ=0n−1(a−κh)a[0]=1
Prove that:
(a+b)[n]=m=0∑n(Cnma[n−m]b[m])
*Cnk=k!(n−k)!n!
And get from this the formula of Newton's Binomial
The base of induction: n=0
(a+b)[0]=m=0∑0(C0ma[0−m]b[m]),⇔1=1
Let's check that the correctness of the statement for n implies the correctness for n+1:
(a+b)[n+1]=m=0∑n+1(Cn+1ma[n+1−m]b[m])
Cnm+Cnm−1=Cn+1m
(a+b)[n+1]==(a+b)[n](a+b−nh)==(m=0∑n(Cnma[n−m]b[m]))(a+b−nh)==(m=0∑n(Cnma[n−m]b[m]))(a+mh−nh+b−mh)==(m=0∑n(Cnma[n−m]b[m]))(a−(n−m)h+b−mh)==(m=0∑n(Cnma[n−m]b[m](a−(n−m)h)))++(m=0∑n(Cnma[n−m]b[m](b−mh)))==(m=0∑n(Cnma[n−m+1]b[m]))++(m=0∑n(Cnma[n−m]b[m+1]))==Cn0a[n+1]b[0]+(m=1∑n(Cnma[n−m+1]b[m]))++(m=1∑n+1(Cn(m−1)a[n−(m−1)]b[(m−1)+1]))=
*Cn0=Cn+k0Cnn=Cn+kn+k
=Cn+10a[n+1]b[0]+(m=1∑n(Cnma[n−m+1]b[m]))++(m=1∑n(Cnm−1a[n−m+1]b[m]))++Cnna[0]b[n+1]==Cn+10a[n+1]b[0]++(m=1∑n((Cnm+Cnm−1)a[n−m+1]b[m]))++Cn+1n+1a[0]b[n+1]==m=0∑n+1(Cn+1ma[n+1−m]b[m])
Which was exactly what had to be proved.
If h=0 α[n]=αn , ⇒
(a+b)n=κ=0∑n(Cnκan−κbκ).