CBSE Class 11 Maths Notes: Binomial Theorem – Problems Related To Series Of Binomial Coefficients

Problem related to series of binomial coefficients in which each term is a product of two binomial coefficients.

Solution Process:

Case I: If difference of the lower suffixes of binomial coefficients in each term is same.

i.e.

C₁C₃ + C₂C₄ + C₃C₅ + …

Here 3 – 1 = 4 – 2 = 5 – 3 = … = 2

Case I: If each term of series is positive then

(1 + x)ⁿ = C₀ + C₁x + C₂x² + …. + C_nxⁿ                                                      …(i)

Interchanging 1 and x,

(x + 1)ⁿ = C₀xⁿ + C₁x^n–1 + C₂x^n–2 + …+ C_n                                                 …(ii)

Then multiplying (i) and (ii) and equate the coefficient of suitable power of x on both sides

Or

Illustration 1:            If I is integral part of (2 + )ⁿ and f is fraction part of (2 + )ⁿ, then prove that (I + f) (1 –f) = 1. Also prove that I is an odd Integer. 

Solution:                   (2 + √3) ⁿ = I + f where I is an integer and 0  ≤ f < 1

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

Since (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 I + f + f ‘ = Even

Now O £f < 1 and 0 < f ‘ < 1                                                ……(1)

Also I + f + f ‘ = Even integer

f + f ‘ = integer                                                                       ……(2)

(1) and (2) imply that f + f ‘ = 1 ( since  0 < f + f ‘ < 2)

⇒ I is odd and f ‘ = 1 – f ⇒ (I + f) (1 – f) = 1.

CBSE Class 11 Maths Binomial Theorem All Topic Notes CBSE Class 11 Maths All Chapters Notes