  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 + …. + Cnxn                                                      …(i)

Interchanging 1 and x,

(x + 1)n = C0xn + C1xn–1 + C2xn–2 + …+ Cn                                                 …(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)= [2n – C1.2n – 1.√3 + C22n – 2.. (√3)– …]

+ [2n  + C1.2n – 1.√3 + C22n -2..(√3)2 – …]

= 2[2n + C2.2n – 2.3+C42n – 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.