1.      Divisibility problems:

        Let (1 + x)ⁿ = 1 + ⁿC₁x + ⁿC₂x² +...+ ⁿC_nxⁿ.

        In any divisibility problem, we have to identify x and n. The number by which division is to be made can be, x, x² or x³, but the number in the base is always expressed in form of (1+x).

Illustration:

        Find the remainder when 7¹⁰³ is divided by 24.

Solution:

        7¹⁰³ = 7(50 - 1)⁵¹ = 7(50⁵¹ - ⁵¹C₁ 50⁵⁰ + ⁵¹C₂ 50⁴⁹ - ... - 1)

        = 7(50⁵¹ - ⁵¹C₁50⁵⁰ +...+ ⁵¹C₅₀50) - 7 - 18 + 18

        = 7(50⁵¹ - ⁵¹C₁50⁵⁰ +...+ ⁵¹C₅₀50) - 25 + 18

        => remainder is 18.

2.     Questions involving the greatest integer function.

        The questions generally involves working with binomial expression on surds.

Illustration:

        If I is the integral part and f is the fraction part of (2 + √3 )ⁿ then prove that (l + f)(1- f) = 1. Also prove that I is an odd integer.

Solution:

        (2 + √3)ⁿ, l + f where I is an integer and 0 , f < 1. We have to show that I is odd and that (l + f)(1 - f) = 1

Here note that (2 + √3)ⁿ (2 + √3)ⁿ =(4 - 3)ⁿ = 1

.·.(2 + √3)ⁿ(2 + √3)ⁿ=1. It is thus required to prove that (2 + √3)ⁿ=1-f

But, (2 + √3)ⁿ + (2 + √3)ⁿ = [2ⁿ - C₁.2^n-1.√3 + C₂2^n-2.(√3)² - ...]

+ [2ⁿ - C₁.2^n-1.√3 + C₂2^n-2.(√3)² - ...]

        =2[2ⁿ - C₂.2^n-2.3 + C₄2^n-4.3² - ...] = even integer

        .·.  Now 0 < (2 - √3) < 1

        .·.  0 < (2 - √3)ⁿ < 1

        .·.  If (2 - √3)n = f' then l + f + f' = Even

        Now ) < f < 1 and 0 < f' < 1                         ... (1)

        .·.  f + f' = integer

        (1) and (2) imply that f + f' = 1(·.· 0 < f + f' < 2)

        .·.  I is odd and f' = 1 - f

        => (I + f)(1 - f) = 1.

Binomial expression with non- positive exponent

        The characteristics discussed hitherto were confined to positive integer n. If n takes any other value, then binomial theorem is written as:

        Say, we have to find out (x+a)ⁿ n I⁺

        If a > 1 then it is written as aⁿ (a + (x/a))ⁿ. This is expanded as

        aⁿ (1+(x/n))ⁿ = aⁿ[1+n(x/a) + (n(n-1)/a) (x/a)² +  (n(n-1)(n-2))/3! (x/a)³ + ......∞]

        Since n is not positive integer therefore the series on the right hand side will converge only for |x/a| < 1. Moreover, there are infinite terms in the expansion contrary to the binomial expansion for a positive integer n.

Multinomial Expression

        If such a case arises, then it is not called Binomial Expansion, it is called Multinomial Expansion. If n ε N, then the general term of the multinomial expansion (x¹ + x² + x³ +...+ x^k)ⁿ is (n!/a₁!a₂!...a_k!) (x₁^a1,x₂^a2,..…x_k^ak ), where a₁ + a₂ + a₃ +......+ a_k = n and a < a_i < n, I = 1, 2, 3, ...... k and the total number of terms in the expansion is ^n+k-1C_n-1.