**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.

C_{1}C_{3} + C_{2}C_{4} + C_{3}C_{5} + …

Here 3 – 1 = 4 – 2 = 5 – 3 = … = 2

**Case I:** If each term of series is positive then

(1 + x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + …. + C_{n}x^{n} …(i)

Interchanging 1 and x,

(x + 1)^{n} = C_{0}x^{n} + C_{1}x^{n–1} + C_{2}x^{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 }= [2^{n} – C_{1}.2^{n – 1}.√3 + C_{2}2^{n – 2.}. (√3)^{2 }– …]

+ [2^{n} + C_{1}.2^{n – 1}.√3 + C_{2}2^{n -2.}.(√3)^{2} – …]

= 2[2^{n} + C_{2}.2^{n – 2}.3+C_{4}2^{n – 4}.3^{2} + …] = 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.

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