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Binomial Theorem - Some special cases in Binomial Expansion

Grade 11CBSE

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

🔑Concepts

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The general expansion of (a+b)n(a + b)^n is given by ∑r=0nnCran−rbr\sum_{r=0}^{n} {}^nC_r a^{n-r} b^r. Special cases arise by substituting specific values for aa and bb.

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Expansion of (x−y)n(x - y)^n: This is obtained by replacing bb with −y-y in the general formula. The terms have alternating signs: (x−y)n=nC0xn−nC1xn−1y+nC2xn−2y2−⋯+(−1)nnCnyn(x - y)^n = {}^nC_0 x^n - {}^nC_1 x^{n-1}y + {}^nC_2 x^{n-2}y^2 - \dots + (-1)^n {}^nC_n y^n.

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Expansion of (1+x)n(1 + x)^n: By putting a=1a = 1 and b=xb = x, we get (1+x)n=nC0+nC1x+nC2x2+⋯+nCnxn(1 + x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + \dots + {}^nC_n x^n. The general term is Tr+1=nCrxrT_{r+1} = {}^nC_r x^r.

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Expansion of (1−x)n(1 - x)^n: By putting a=1a = 1 and b=−xb = -x, we get (1−x)n=1−nC1x+nC2x2−nC3x3+⋯+(−1)nnCnxn(1 - x)^n = 1 - {}^nC_1 x + {}^nC_2 x^2 - {}^nC_3 x^3 + \dots + (-1)^n {}^nC_n x^n.

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Sum and Difference of expansions: (x+a)n+(x−a)n(x+a)^n + (x-a)^n results in 2×2 \times (sum of terms at odd positions), while (x+a)n−(x−a)n(x+a)^n - (x-a)^n results in 2×2 \times (sum of terms at even positions).

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The number of terms in the expansion of (a+b)n(a+b)^n is always n+1n+1.

📐Formulae

(x−y)n=∑r=0n(−1)rnCrxn−ryr(x - y)^n = \sum_{r=0}^{n} (-1)^r {}^nC_r x^{n-r} y^r

(1+x)n=1+nx+n(n−1)2!x2+⋯+xn(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots + x^n

(x+a)n+(x−a)n=2[nC0xn+nC2xn−2a2+nC4xn−4a4+… ](x + a)^n + (x - a)^n = 2 [ {}^nC_0 x^n + {}^nC_2 x^{n-2} a^2 + {}^nC_4 x^{n-4} a^4 + \dots ]

(x+a)n−(x−a)n=2[nC1xn−1a+nC3xn−3a3+nC5xn−5a5+… ](x + a)^n - (x - a)^n = 2 [ {}^nC_1 x^{n-1} a + {}^nC_3 x^{n-3} a^3 + {}^nC_5 x^{n-5} a^5 + \dots ]

Tr+1=nCrxr for the expansion of (1+x)nT_{r+1} = {}^nC_r x^r \text{ for the expansion of } (1+x)^n

💡Examples

Problem 1:

Expand (1−2x)5(1 - 2x)^5 using the binomial theorem.

Solution:

Using the formula for (1−x)n(1 - x)^n where n=5n=5 and the variable is 2x2x: (1−2x)5=5C0(1)5−5C1(1)4(2x)+5C2(1)3(2x)2−5C3(1)2(2x)3+5C4(1)1(2x)4−5C5(2x)5(1 - 2x)^5 = {}^5C_0(1)^5 - {}^5C_1(1)^4(2x) + {}^5C_2(1)^3(2x)^2 - {}^5C_3(1)^2(2x)^3 + {}^5C_4(1)^1(2x)^4 - {}^5C_5(2x)^5 =1−5(2x)+10(4x2)−10(8x3)+5(16x4)−1(32x5)= 1 - 5(2x) + 10(4x^2) - 10(8x^3) + 5(16x^4) - 1(32x^5) =1−10x+40x2−80x3+80x4−32x5= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5

Explanation:

We applied the special case (1−x)n(1-x)^n where xx is replaced by 2x2x. The signs alternate because of the negative sign in the binomial.

Problem 2:

Find the value of (2+1)6+(2−1)6(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6.

Solution:

Let x=2x = \sqrt{2} and a=1a = 1. We use the formula (x+a)n+(x−a)n=2[nC0xn+nC2xn−2a2+nC4xn−4a4+nC6xn−6a6](x+a)^n + (x-a)^n = 2[ {}^nC_0 x^n + {}^nC_2 x^{n-2} a^2 + {}^nC_4 x^{n-4} a^4 + {}^nC_6 x^{n-6} a^6 ]. For n=6n=6: 2[6C0(2)6+6C2(2)4(1)2+6C4(2)2(1)4+6C6(1)6]2 [ {}^6C_0 (\sqrt{2})^6 + {}^6C_2 (\sqrt{2})^4 (1)^2 + {}^6C_4 (\sqrt{2})^2 (1)^4 + {}^6C_6 (1)^6 ] =2[1(8)+15(4)+15(2)+1(1)]= 2 [ 1(8) + 15(4) + 15(2) + 1(1) ] =2[8+60+30+1]=2[99]=198= 2 [ 8 + 60 + 30 + 1 ] = 2 [ 99 ] = 198

Explanation:

When adding (x+a)n(x+a)^n and (x−a)n(x-a)^n, the terms containing odd powers of aa cancel out, leaving twice the sum of terms containing even powers of aa.

Problem 3:

Find the middle term in the expansion of (1+x)10(1 + x)^{10}.

Solution:

Here n=10n = 10, which is even. The number of terms is n+1=11n + 1 = 11. The middle term is the (102+1)th(\frac{10}{2} + 1)^{th} term, which is the 6th6^{th} term (T6T_6). Using Tr+1=nCrxrT_{r+1} = {}^nC_r x^r for (1+x)n(1+x)^n: T6=T5+1=10C5x5T_6 = T_{5+1} = {}^{10}C_5 x^5 10C5=10×9×8×7×65×4×3×2×1=252{}^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 Middle term =252x5= 252x^5.

Explanation:

For an even nn, there is only one middle term located at position n2+1\frac{n}{2} + 1.

Some special cases in Binomial Expansion Class 11 Notes & Examples