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Binomial Theorem - Simple Applications

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

The Binomial Theorem provides a formula for the expansion of (a+b)n(a+b)^n for any positive integer nn. Visually, this expansion follows the structure of Pascal's Triangle, a symmetric pyramid where each number is the sum of the two directly above it, representing the binomial coefficients nCr{^nC_r}.

The total number of terms in the expansion of (a+b)n(a+b)^n is always n+1n+1. For example, a squared binomial (a+b)2(a+b)^2 has 3 terms, and a cubed binomial (a+b)3(a+b)^3 has 4 terms.

The General Term, denoted as Tr+1T_{r+1}, is a single expression that represents any term in the expansion. It is given by Tr+1=nCranrbrT_{r+1} = {^nC_r} a^{n-r} b^r. This formula is the primary tool for finding a specific power of xx or a term at a specific position without writing the full expansion.

The Middle Term depends on the value of nn. If nn is even, the expansion has an odd number of terms, resulting in one middle term at position (n2+1)(\frac{n}{2} + 1). If nn is odd, the expansion has an even number of terms, resulting in two middle terms at positions (n+12)(\frac{n+1}{2}) and (n+32)(\frac{n+3}{2}). Visually, these terms correspond to the peak values in the distribution of coefficients.

The term independent of xx (the constant term) occurs when the total exponent of xx in the general term equals zero. To find it, one must aggregate all powers of xx from both parts of the binomial and solve for rr such that the resulting exponent is 0.

The sum of all binomial coefficients in the expansion of (1+x)n(1+x)^n is 2n2^n. This is derived by substituting x=1x=1 into the identity. Visually, this is equivalent to summing all the values in the nthn^{th} row of Pascal's Triangle.

Binomial expansions are used for numerical approximations. For example, (1.01)10(1.01)^{10} can be written as (1+0.01)10(1 + 0.01)^{10}. By calculating only the first few terms of the expansion, one can obtain a highly accurate approximation because subsequent terms involve powers of 0.010.01 which become insignificantly small.

📐Formulae

(a+b)n=nC0an+nC1an1b+nC2an2b2+...+nCnbn(a+b)^n = {^nC_0} a^n + {^nC_1} a^{n-1}b + {^nC_2} a^{n-2}b^2 + ... + {^nC_n} b^n

(a+b)n=r=0nnCranrbr(a+b)^n = \sum_{r=0}^{n} {^nC_r} a^{n-r} b^r

nCr=n!r!(nr)!{^nC_r} = \frac{n!}{r!(n-r)!}

Tr+1=nCranrbrT_{r+1} = {^nC_r} a^{n-r} b^r

If n is even, Middle Term =Tn2+1\text{If } n \text{ is even, Middle Term } = T_{\frac{n}{2}+1}

If n is odd, Middle Terms =Tn+12 and Tn+32\text{If } n \text{ is odd, Middle Terms } = T_{\frac{n+1}{2}} \text{ and } T_{\frac{n+3}{2}}

Sum of coefficients: r=0nnCr=2n\text{Sum of coefficients: } \sum_{r=0}^{n} {^nC_r} = 2^n

💡Examples

Problem 1:

Find the coefficient of x6x^6 in the expansion of (3x213x)9(3x^2 - \frac{1}{3x})^9.

Solution:

  1. Write the general term Tr+1=9Cr(3x2)9r(13x)rT_{r+1} = {^9C_r} (3x^2)^{9-r} (-\frac{1}{3x})^r.
  2. Separate the coefficients and the xx variables: Tr+1=9Cr(3)9r(x2)9r(13)r(x1)rT_{r+1} = {^9C_r} (3)^{9-r} (x^2)^{9-r} (-\frac{1}{3})^r (x^{-1})^r.
  3. Combine the powers of xx: Tr+1=9Cr(3)9r(13)rx182rr=9Cr(3)9r(13)rx183rT_{r+1} = {^9C_r} (3)^{9-r} (-\frac{1}{3})^r x^{18-2r-r} = {^9C_r} (3)^{9-r} (-\frac{1}{3})^r x^{18-3r}.
  4. To find the coefficient of x6x^6, set the exponent 183r=618-3r = 6.
  5. Solve for rr: 3r=12    r=43r = 12 \implies r = 4.
  6. Substitute r=4r=4 back into the coefficient part: 9C4(3)94(13)4=9C4(3)5(134)=9C4×3{^9C_4} (3)^{9-4} (-\frac{1}{3})^4 = {^9C_4} (3)^5 (\frac{1}{3^4}) = {^9C_4} \times 3.
  7. Calculate 9C4=9×8×7×64×3×2×1=126{^9C_4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126.
  8. Final Coefficient =126×3=378= 126 \times 3 = 378.

Explanation:

We use the General Term formula to express the r+1r+1-th term. By grouping the exponents of xx and setting them equal to 6, we determine that the 5th term (r=4r=4) contains x6x^6. Then we calculate the numerical value.

Problem 2:

Find the term independent of xx in the expansion of (x2x2)12(x - \frac{2}{x^2})^{12}.

Solution:

  1. Write the general term Tr+1=12Cr(x)12r(2x2)rT_{r+1} = {^{12}C_r} (x)^{12-r} (-\frac{2}{x^2})^r.
  2. Simplify the powers of xx: Tr+1=12Crx12r(2)rx2r=12Cr(2)rx123rT_{r+1} = {^{12}C_r} x^{12-r} (-2)^r x^{-2r} = {^{12}C_r} (-2)^r x^{12-3r}.
  3. For the term independent of xx, the exponent of xx must be zero: 123r=012-3r = 0.
  4. Solve for rr: 3r=12    r=43r = 12 \implies r = 4.
  5. The term independent of xx is T4+1=T5T_{4+1} = T_5.
  6. Calculate the value: T5=12C4(2)4=495×16=7920T_5 = {^{12}C_4} (-2)^4 = 495 \times 16 = 7920.

Explanation:

The term independent of xx is the constant term where the power of xx is x0x^0. We find rr by setting the combined exponent of xx to zero and then compute the value of that specific term.