Subtopic: Newton's Binomial theorem (formula)

= (3)

Let us consider special cases of this formula for n = 2 and n = 3.

= (a + x)*(a + x) = aa + ax + xa + xx; (3)

= +(a + x) = (aa + ax + xa + xx)*(a + x)= = aaa + axa + xaa + xxa + aax + axx + xax + xxx; (4)

From equalities (3) and (4), we can see that their right parts contain:

- all 2-permutations made of a and x in (3);

- all 3-permutations made of a and x in (4);

The same is true in a general case:

(a + x)*(a + x)*...*(a + n)

n times

When we remove the brackets we obtain all possible n-permutations made of letters ‘x’ and ‘a’.

Collect like terms. They contain an equal number of letters ‘x’ (so of letters ‘a’ also).

Let us find how many terms contain k letters 'x', hence n-k letters 'a'.

By the formula of the number of permutations with a specification, the number of such terms is equal to:

P(k, n-k) = = ;

From here it follows that after collecting like terms, the term x^k * a^n-k enters the sum with the coefficient . So we have proved that =

H/a. Using the Newton's binomial formula present the expression as a sum of terms.