Suppose k power again

Finally, when k+ 1 is multiplied by k power (a+b), the problem is solved by polynomial multiplication.

Example: prove that when n= 1, the left side = (a+b)1= a+b.

Right = c 01a+c11b = a+b

Left = right

Assuming that when n = k, the equation holds,

That is, (a+b) n = c0nan+c1na (n-1) b10? Ten Crn a(n-r)br ten? Ten Cnn bn was established;

Then when n=k+ 1, (a+b) (n+1) = (a+b) n * (a+b) = [c0nan+c1na (n-1) b/kl. Ten Crn a(n-r)br ten? Ten Cnn bn]*(a+b)

=[c0nan+c 1na(n- 1)b 10？ Ten Crn a(n-r)br ten? Ten cnnnn] * a+[c0nan+c1na (n-1) b10? Ten Crn a(n-r)br ten? ten CNN nn]=[c0na(n+ 1)+c 1 nanb ten？ Ten Crn a(n-r+ 1)br Ten？ Ten cnnabn]+[c0nanb+c1na (n-1) B2 ten? Ten Crn a(n-r)b(r+ 1) ten？ Cnn b(n+ 1)]

= c0na (n+1)+(c0n+c1n) anb ten? Ten(c(r- 1)n+CrN)a(n-r+ 1)br Ten？ Ten (c (n-1) n+CNN) ABN+CNNB (n+1)]

= C0(n+ 1)a(n+ 1)+c 1(n+ 1)anb+C2(n+ 1)a(n- 1)B2+？ +Cr(n+ 1)a(n-r+ 1)br+？ +C(n+ 1)(n+ 1)b(n+ 1)

∴ When n=k+ 1, the equation also holds;

So for any positive integer, the equation holds.