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Algebra - Binomial Theorem

Grade 11ICSE

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

🔑Concepts

Binomial Expression and Expansion: A binomial is an algebraic expression containing two terms, such as (a+b)(a + b). The Binomial Theorem provides a formula for expanding any positive integer power of a binomial, (a+b)n(a + b)^n. In the expansion, the powers of the first term 'aa' decrease from nn to 00, while the powers of the second term 'bb' increase from 00 to nn.

Pascal's Triangle (Visual Concept): This is a geometric arrangement of binomial coefficients. It looks like a triangle where the number at the apex is 11. Each number below is the sum of the two numbers directly above it. Visually, the triangle is perfectly symmetric; the coefficients on the left side are identical to those on the right, which corresponds to the property nCr=nCnr^{n}C_r = ^{n}C_{n-r}.

General Term (Tr+1T_{r+1}): The (r+1)th(r+1)^{th} term in the expansion of (a+b)n(a + b)^n is denoted by Tr+1T_{r+1}. This formula allows us to find any specific term in the expansion without writing out the whole series. Note that to find the kthk^{th} term, we substitute r=k1r = k-1.

Total Number of Terms: The number of terms in the expansion of (a+b)n(a + b)^n is always n+1n + 1. For example, (a+b)2(a + b)^2 expands to a2+2ab+b2a^2 + 2ab + b^2, which contains 33 terms. This is a linear progression where the count of terms is always one more than the index of the power.

Middle Term(s): The position of the middle term depends on whether nn is even or odd. If nn is even, the expansion has an odd number of terms, resulting in one middle term at the (n2+1)th(\frac{n}{2} + 1)^{th} position. If nn is odd, the expansion has an even number of terms, resulting in two middle terms at the (n+12)th(\frac{n+1}{2})^{th} and (n+32)th(\frac{n+3}{2})^{th} positions. Visually, these represent the central values in a row of Pascal's Triangle.

Properties of Coefficients: The sum of all binomial coefficients in the expansion of (1+x)n(1 + x)^n is 2n2^n (obtained by setting x=1x = 1). Furthermore, the sum of coefficients of odd terms is equal to the sum of coefficients of even terms, both being equal to 2n12^{n-1}.

Index Rules: In every term of the expansion of (a+b)n(a + b)^n, the sum of the indices (exponents) of aa and bb is always equal to nn. For instance, in the expansion of (x+y)5(x + y)^5, a term like x2y3x^2y^3 has a total index sum of 2+3=52 + 3 = 5.

📐Formulae

General Expansion: (a+b)n=nC0an+nC1an1b+nC2an2b2++nCnbn(a + b)^n = ^{n}C_0 a^n + ^{n}C_1 a^{n-1}b + ^{n}C_2 a^{n-2}b^2 + \dots + ^{n}C_n b^n

Sigma Notation: (a+b)n=r=0nnCranrbr(a + b)^n = \sum_{r=0}^{n} {^{n}C_r} a^{n-r} b^r

Binomial Coefficient: nCr=n!r!(nr)!^{n}C_r = \frac{n!}{r!(n-r)!}

General Term: Tr+1=nCranrbrT_{r+1} = ^{n}C_r a^{n-r} b^r

Middle Term (n is even): Tn2+1T_{\frac{n}{2} + 1}

Middle Terms (n is odd): Tn+12T_{\frac{n+1}{2}} and Tn+32T_{\frac{n+3}{2}}

Expansion of (1+x)n(1+x)^n: 1+nx+n(n1)2!x2+n(n1)(n2)3!x3++xn1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots + x^n

💡Examples

Problem 1:

Find the 5th5^{th} term in the expansion of (x22x)9(x^2 - \frac{2}{x})^9.

Solution:

  1. Identify the parameters: Here a=x2a = x^2, b=2xb = -\frac{2}{x}, and n=9n = 9.\n2. For the 5th5^{th} term (T5T_5), we use r=4r = 4 in the general term formula Tr+1=nCranrbrT_{r+1} = ^{n}C_r a^{n-r} b^r.\n3. Substitute the values: T5=9C4(x2)94(2x)4T_5 = ^{9}C_4 (x^2)^{9-4} (-\frac{2}{x})^4.\n4. Calculate 9C4^{9}C_4: 9C4=9×8×7×64×3×2×1=126^{9}C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126.\n5. Simplify the expression: T5=126(x2)5(16x4)=126x1016x4T_5 = 126 \cdot (x^2)^5 \cdot (\frac{16}{x^4}) = 126 \cdot x^{10} \cdot \frac{16}{x^4}.\n6. Final calculation: T5=126×16x6=2016x6T_5 = 126 \times 16 \cdot x^6 = 2016x^6.

Explanation:

We use the general term formula Tr+1T_{r+1}. Since we need the 5th term, rr must be 4. We carefully include the negative sign with the second term bb and simplify the powers of xx using exponent laws.

Problem 2:

Find the term independent of xx in the expansion of (2x+13x2)9(2x + \frac{1}{3x^2})^9.

Solution:

  1. Write the general term Tr+1=9Cr(2x)9r(13x2)rT_{r+1} = ^{9}C_r (2x)^{9-r} (\frac{1}{3x^2})^r.\n2. Separate the constants and variables: Tr+1=9Cr(2)9r(x)9r(13)r(x2)rT_{r+1} = ^{9}C_r (2)^{9-r} (x)^{9-r} (\frac{1}{3})^r (x^{-2})^r.\n3. Combine the powers of xx: Tr+1=9Cr29r3rx9r2r=9Cr29r3rx93rT_{r+1} = ^{9}C_r 2^{9-r} 3^{-r} x^{9-r-2r} = ^{9}C_r 2^{9-r} 3^{-r} x^{9-3r}.\n4. For the term to be independent of xx, the exponent of xx must be zero: 93r=0    3r=9    r=39 - 3r = 0 \implies 3r = 9 \implies r = 3.\n5. Substitute r=3r=3 into the constant part: T4=9C329333=9C326127T_4 = ^{9}C_3 2^{9-3} 3^{-3} = ^{9}C_3 2^6 \frac{1}{27}.\n6. Calculate 9C3=9×8×73×2×1=84^{9}C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84.\n7. Result: T4=84×64×127=28×649=17929T_4 = 84 \times 64 \times \frac{1}{27} = \frac{28 \times 64}{9} = \frac{1792}{9}.

Explanation:

A term 'independent of xx' means the power of xx in that term is 00. We find the general term, group all powers of xx together, set that total exponent to 00 to find the value of rr, and then calculate the coefficient for that rr.