Properties of Binomial Expansion

 

Binomial theorem or the Binomial expansion is an important component of IIT JEE Mathematics syllabus. It is neither very simple nor extremely difficult and fetches some direct questions in various competitions. We first consider some of the important concepts related with the binomial theorem properties and then proceed towards some questions based on them.

Binomial theorem tells us how to expand some integral power of a binomial expansion in the form of series. If x, y ∈ R and n ∈ N, we have

(x+y)n = nC0 xn + nC1 xn-1y + nC2 xn-2 y2 +…… … .. + nCr xn-r yr + ….. + nCn yn = Σ nCr xn-ryr

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Some Key Points: 

• The number of terms in the expansion is (n+1) which is one more than the index.

• The sum of the indices in x and y is 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 

=>  nCr = nCn-r = n!/r!(n-r)!

 

• The general term in the expansion of (a + x)ⁿ is (r + 1)^th term given as

Tr+1 = nCran-r + xr.

•  Some of the standard properties of binomial coefficients which should be remembered are:

C₀ + C₁ + C₂ + ….. + C_n = 2ⁿ

C₀ + C₂ + C₄ + ….. = C₁ + C₃ + C₅ + ….. = = 2^n-1

C₀² + C₁² + C₂² + ….. + C_n² = ²ⁿ C_n = (2n!)/ n!n!

• 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.

• 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.

• Another result that is applied in questions is ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r.

• We can also replace ^mC₀ by ^m+1C₀ because numerical value of both is same i.e. 1. Similarly we can replace ^mC_m by ^m+1C_m+1.

We now proceed towards some of the illustrations:

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 is

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

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

Illustration: If (1+ax)ⁿ = 1+8x + 24x² + …. , then find the value of a and n. (1983) 

Solution: It is given in the question that

(1+ax)ⁿ = 1 + 8x + 24x² + ….

Hence, 1+ anx + n(n-1)/2! a²x² + ….. = 1 + 8x + 24x² + ….

This gives an = 8 and a²n(n-1)/ 2 = 24.

Hence, 8(8-a) = 48.

This implies 8-a = 6.

Hence, a = 2 and n = 4.

Illustration: Let n be a positive integer. If the coefficients of 2^nd, 3^rd and 4^th terms in the expansion of (1+x)ⁿ are in A.P. then what is the value of n? (1994) 

Solution: Let the coefficients of 2^nd, 3^rd and 4^th terms in the expansion of (1+x)ⁿ be ⁿC₁, ⁿC₂, ⁿC₃.

According to the given condition,

2 (ⁿC₂) = ⁿC₁ + ⁿC₃

2. n(n-1)/ 1.2 = n + n(n-1)(n-2)/ 1.2.3

Hence, n-1 = 1+ (n-1)(n-2)/6

So, n-1 = 1+ (n-1)(n-2)/6

This gives, n-1 = 1+ (n²-3n+2)/6

Hence, 6n-6 = 6+ n² -3n +2

This gives n²-9n+14 = 0

So, (n-2)(n-7) = 0

Hence, either n = 2 or n = 7.

But ⁿC₃ is true for n ≥ 3, hence this gives the value of n as 7.

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