1.   There are (n + 1) terms in the expansion of (a + b)ⁿ, the first and the last term being aⁿ and bⁿ respectively. If ⁿC_x = ⁿC_y, then either x = y or x + y = n 

       =>  ⁿC_r = ⁿC_n-r = n!/r!(n-r)! .

2.   The general term in the expansion of (a + x)ⁿ is (r + 1)^th term given as T_r+1 = ⁿC_ra^n-r + x^r.

        Similarly the general term in the expansion of (x + a)ⁿ is given as T_r+1 = ⁿC_r x^n-r a^r. The terms are considered from the beginning.

3.    The binomial coefficient in the expansion of (a + x)ⁿ which are equidistant from the beginning and the end are equal i.e. ⁿC_r = ⁿC_n-r.

Note:  Here we are using ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r this concept will be discussed later in this chapter. Also, we have replace ^mC₀ by ^m+1C₀ because numerical value of both is same i.e. 1. Similarly we replace ^mC_m by ^m+1C_m+1.

Illustration:

        Find the expansion of (a-x)ⁿ.

Solution: We know that

       (a + x)ⁿ = aⁿ + na^(n-1) x +   (n(n-1)/2!) a^(n-2) x² +...+ xⁿ.

       putting x = -x in the above expansion, we get,

       (a-x)ⁿ = [a + (-x)]ⁿ = aⁿ + na^(n-1) (-x) + (n(n-1)/2!) a^(n-2) (-x)² +......+ (-x)ⁿ.

        = aⁿ - na^(n-1) x + (n(n-1)/2!) a^(n-2) x² -......+ (-1)ⁿ xⁿ.

Illustration:

        Find the value of  (a+√(a² -1))⁷ +(a-√(a² -1))⁷.

Solution:

        Here, we have to find the sum of two expansions whose terms are numerically the same, but in the second expansion the second, fourth, sixth and eight terms are negative, and therefore cancel the corresponding terms of the first expansion. Hence, the given expression

        = 2 {a⁷ + 21 a⁵ (a² - 1)+ 35 a³(a² - 1)² + 7a (a² - 1)³}

        = 2a (64 a⁶ - 112 a⁴ + 56 a² - 7)