JEE Main 2024 Question Paper Solution Discussion Live JEE Main 2024 Question Paper Solution Discussion Live

Results for Specific Types of Binomial Expansions

The concept of binomial theorem dates back to the 4th century BC. It was introduced by a Greek Mathematician named Euclid. Al-Karaji was the person whose contribution was more to the topic of the binomial theorem. He used the idea of a triangular pattern to explain the coefficients of the binomial theorem. He was the writer of the proof of the binomial theorem and explained Pascal’s triangle. The Binomial Expansion elucidates the numerical expansion of the exponentiation of an expression that is binomial in nature. The expansion of the binomial theorem is given below:

(1 + x)n = nC0 + nC1 x + nC2 x2 + ……. + nCn xn

Features of binomial expansion are as follows:

  • There exists (n + 1) terms in the expansion of (x + y)n.
  • The power of x begins at n and decreases until it reduces to 0.
  • The power of y begins at 0 and increases until it reaches n.
  • In a binomial expansion, the coefficient of binomial expression is equidistant in nature till the end.
  • The binomial coefficients can be represented using Pascal’s triangle.

Results for Specific Types of Binomial Expansions

(a + x)n ± or (✕) (a – x)n type

1] (a + x)n = nC0 an + nC1 an-1 x + nC2 an-2 x2 + ……. + nCn a0 xn

2] (a – x)n = nC0 annC1 an-1 x + nC2 an-2 x2 + ……. + nCn a0 (- x)n

3] Sum to n odd terms = T1 + T3 + … are odd terms = A (let)

Sum to n even terms = T2 + T4 + ……. are even terms = B (let)

4] (a + x)n = nC0 an + nC1 an-1 x + nC2 an-2 x2 + ……. + nCn a0 xn = A + B

(a – x)n = nC0 annC1 an-1 x + nC2 an-2 x2 + ……. + nCn a0 (- x)n = A – B

5] (a + x)n = A + B

(a – x)n = A – B

(a + x)n + (a – x)n = A + B + A – B = 2A

= 2 * odd terms

= 2 [T1 + T3 + T5 + ….. ]

6] (a + x)n – (a – x)n = (A + B) – (A – B) = 2B

= 2 * even terms

= 2 [T2 + T4 + T6 ……. ]

7] (a2 – x2)n = (a + x)n . (a – x)n

= (A + B) (A – B)

= A2 – B2

Example 1: The value of (30C0) (30C10) − (30C1) (30C11) + (30C2) (30C12) + …… + (30C20) (30C30)

A) 60C20

B) 30C10

C) 60C30

D) 40C30

Solution: B

(1 − x)30 = 30C0 x030C1 x1 + 30C2 x2 + …… + (−1)30 30C30 x30 …. (i)

(x + 1)30 = 30C0 x30 + 30C1 x29 + 30C2 x28 + …… + 30C10 x20 + …. + 30C30 x0 ….(ii)

Multiplying (i) and (ii) and equating the coefficient of x20 on both sides, we get the required sum

= coefficient of x20 in (1 – x2)30

= 30C10

Example 2: The coefficient of the term independent of x in the expansion of

\(\begin{array}{l}(1+x+2{{x}^{3}}){{\left( \frac{3}{2}{{x}^{2}}-\frac{1}{3x} \right)}^{9}}\end{array} \)
is

A) 1/ 3

B) 19 / 54

C) 17 / 54

D) 1 / 4

Solution: C

The general term in the expansion of

\(\begin{array}{l}{{\left( \frac{3}{2}{{x}^{2}}-\frac{1}{3x} \right)}^{9}} is {{T}_{r+1}}={{\,}^{9}}{{C}_{r}}{{\left( \frac{3}{2}{{x}^{2}} \right)}^{9-r}}{{\left( -\frac{1}{3x} \right)}^{r}}\\ ={{\,}^{9}}{{C}_{r}}{{\left( \frac{3}{2} \right)}^{9-r}}\left( -\frac{1}{3} \right){{x}^{18-3r}}……(i) \\ \text { Now, the coefficient of the term independent of x in the expansion of } (1+x+2{{x}^{3}})\,\,{{\left( \frac{3}{2}{{x}^{2}}-\frac{1}{3x} \right)}^{9}} ……(ii) \\ = \text { Sum of the coefficient of the terms } {{x}^{0}},{{x}^{-1}}\ \text{and}\ {{x}^{-3}} in {{\left( \frac{3}{2}{{x}^{2}}-\frac{1}{3x} \right)}^{9}}\end{array} \)

For x0 in (i) above, 18 – 3r = 0 ⇒ r = 6

For x-1 in (i) above, there exists no value of r and hence no such term exists.

For x-3 in (i), 18 – 3r = -3 ⇒ r = 7

For the term independent of x, in (ii), the coefficient

\(\begin{array}{l}=1\times {{\,}^{9}}{{C}_{6}}{{(-1)}^{6}}{{\left( \frac{3}{2} \right)}^{9-6}}{{\left( \frac{1}{3} \right)}^{6}}+2\times {{\,}^{9}}{{C}_{7}}{{(-1)}^{7}}{{\left( \frac{3}{2} \right)}^{9-7}}{{\left( \frac{1}{3} \right)}^{7}}\\ =\frac{9.8.7}{1.2.3}.\frac{{{3}^{3}}}{{{2}^{3}}}.\frac{1}{{{3}^{6}}}+2\frac{9.8}{1.2}(-1)\frac{{{3}^{2}}}{{{2}^{2}}}.\frac{1}{{{3}^{7}}}\\ =\frac{7}{18}-\frac{2}{27}=\frac{17}{54}.\end{array} \)

Results for Specific Types of Binomial Expansions – Video Lesson

JEE Maths

Frequently Asked Questions

Q1

Give the binomial expansion formula for (1 + x)n .

The binomial expansion formula is given by (1 + x)n = nC0 + nC1 x + nC2 x2 + ……. + nCn xn.

Q2

What is the number of terms in binomial expansion (x+y)n?

The number of terms in the binomial expansion (x+y)n is n + 1.

Q3

Where is the binomial theorem used?

Binomial Theorem is used to find roots of equations in higher powers. It is used to find the expansion for the algebraic identities.

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