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Binomial Theorem - Binomial theorem for positive integral indices

Grade 11CBSE

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

🔑Concepts

Definition of Binomial Theorem: The Binomial Theorem for positive integral indices states that for any positive integer nn and any real numbers aa and bb, the expansion of (a+b)n(a+b)^n consists of (n+1)(n+1) terms. The coefficients of these terms are binomial coefficients denoted by nCr^nC_r.

Expansion Pattern and Sum of Indices: In the expansion of (a+b)n(a+b)^n, the power of 'aa' starts at nn and decreases by 1 in each subsequent term until it reaches 0. Simultaneously, the power of 'bb' starts at 0 and increases by 1 until it reaches nn. Visually, if you look at any term, the sum of the exponents of aa and bb is always equal to the original index nn.

Pascal's Triangle: The coefficients of the binomial expansion can be arranged in a geometric, triangular pattern called Pascal's Triangle. It begins with 1 at the top (n=0n=0). Each row starts and ends with 1, and every interior number is the sum of the two numbers directly above it. This provides a clear visual symmetry to the coefficients of the expansion.

The General Term (Tr+1T_{r+1}): Any specific term in the expansion can be located using the general term formula Tr+1=nCranrbrT_{r+1} = ^nC_r a^{n-r} b^r. Here, rr represents the index of the combination and is always one less than the position of the term (e.g., for the 5th term, r=4r=4).

Middle Terms: The number of terms in the expansion is (n+1)(n+1). If nn is even, there is a single middle term at position (n2+1)(\frac{n}{2} + 1). If nn is odd, the number of terms (n+1)(n+1) is even, meaning there are two middle terms at positions (n+12)(\frac{n+1}{2}) and (n+12+1)(\frac{n+1}{2} + 1). This can be visualized as the 'peak' of the distribution of coefficients.

Symmetry of Coefficients: The coefficients in the expansion are symmetrical. The first coefficient nC0^nC_0 is equal to the last nCn^nC_n, the second nC1^nC_1 is equal to the second-to-last nCn1^nC_{n-1}, and so on. This follows the visual symmetry of Pascal's Triangle and the mathematical identity nCr=nCnr^nC_r = ^nC_{n-r}.

📐Formulae

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

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

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

Middle term (if nn is even) = Tfracn2+1T_{\\frac{n}{2}+1}

Middle terms (if nn is odd) = Tfracn+12T_{\\frac{n+1}{2}} and Tfracn+12+1T_{\\frac{n+1}{2}+1}

(1+x)n=nC0+nC1x+nC2x2+dots+nCnxn(1+x)^n = ^nC_0 + ^nC_1 x + ^nC_2 x^2 + \\dots + ^nC_n x^n

Sum of all binomial coefficients: \\sum_{r=0}^{n} ^nC_r = 2^n

💡Examples

Problem 1:

Expand (x+2)4(x + 2)^4 using the Binomial Theorem.

Solution:

Step 1: Identify a=xa = x, b=2b = 2, and n=4n = 4. Total terms = 4+1=54 + 1 = 5. \nStep 2: Apply the formula (a+b)^n = \\sum_{r=0}^n ^nC_r a^{n-r} b^r: \nT1=4C0x420=1cdotx4cdot1=x4T_1 = ^4C_0 x^4 2^0 = 1 \\cdot x^4 \\cdot 1 = x^4 \nT2=4C1x321=4cdotx3cdot2=8x3T_2 = ^4C_1 x^3 2^1 = 4 \\cdot x^3 \\cdot 2 = 8x^3 \nT3=4C2x222=6cdotx2cdot4=24x2T_3 = ^4C_2 x^2 2^2 = 6 \\cdot x^2 \\cdot 4 = 24x^2 \nT4=4C3x123=4cdotxcdot8=32xT_4 = ^4C_3 x^1 2^3 = 4 \\cdot x \\cdot 8 = 32x \nT5=4C4x024=1cdot1cdot16=16T_5 = ^4C_4 x^0 2^4 = 1 \\cdot 1 \\cdot 16 = 16 \nStep 3: Combine the terms: x4+8x3+24x2+32x+16x^4 + 8x^3 + 24x^2 + 32x + 16.

Explanation:

We use the expansion formula term by term, calculating combinations nCr^nC_r and powers of xx and 2 separately before multiplying them together.

Problem 2:

Find the middle term in the expansion of (x2+frac1x)6(x^2 + \\frac{1}{x})^6.

Solution:

Step 1: Identify n=6n=6. Since nn is even, there is one middle term. \nStep 2: The position of the middle term is (fracn2+1)=(frac62+1)=4(\\frac{n}{2} + 1) = (\\frac{6}{2} + 1) = 4. So, we need to find T4T_4. \nStep 3: For T4T_4, r=3r = 3. Use the general term formula Tr+1=nCranrbrT_{r+1} = ^nC_r a^{n-r} b^r with a=x2a = x^2 and b=frac1xb = \\frac{1}{x}. \nT4=6C3(x2)63(frac1x)3T_4 = ^6C_3 (x^2)^{6-3} (\\frac{1}{x})^3 \nT4=20cdot(x2)3cdotfrac1x3T_4 = 20 \\cdot (x^2)^3 \\cdot \\frac{1}{x^3} \nT4=20cdotx6cdotfrac1x3T_4 = 20 \\cdot x^6 \\cdot \\frac{1}{x^3} \nT4=20x3T_4 = 20x^3. \nThe middle term is 20x320x^3.

Explanation:

Since n=6n=6, there are 7 terms in total, making the 4th term the exact middle. We use the general term formula with r=3r=3 to solve for it.