Binomial Coefficients — Properties, sum of coefficients
Binomial Theorem
9
JEE Qs
8%
Hard
60
min
Master the art of deriving various sums by substituting values, differentiating, or integrating the binomial expansion, and recognize common combinatorial identities for efficient problem-solving.
🧮 Key Formulas
✅ Key Points for JEE
- 1Many summation properties of binomial coefficients (e.g., sum of all, alternating sum) can be derived by substituting specific values for variables (e.g., x=1, y=1 or x=1, y=-1) in the general binomial expansion.
- 2Sums involving r * nCr or r^2 * nCr can often be solved by differentiating the binomial expansion (1+x)^n with respect to x, and then substituting x=1.
- 3Sums involving nCr/(r+1) can often be solved by integrating the binomial expansion (1+x)^n with respect to x, and then substituting appropriate limits.
- 4Recognize and apply fundamental combinatorial identities like symmetry (nCr = nC(n-r)), Pascal's Identity (nCr + nC(r-1) = (n+1)Cr), and the identity r * nCr = n * (n-1)C(r-1) for simplification.
- 5Sums involving products of binomial coefficients, like Σ (nCr)(mCk-r), are often related to Vandermonde's Identity; a common case is Σ (nCr)^2 = (2n)Cn, which can be derived by comparing coefficients of x^n in (1+x)^n * (1+x)^n = (1+x)^(2n).
⚠️ Common Mistakes
- ✕Incorrectly applying differentiation or integration techniques without adjusting the summation limits or terms properly.
- ✕Confusing (Σ nCr)^2 with Σ (nCr)^2, leading to incorrect calculations.
- ✕Errors in algebraic manipulation of 'nCr' or not recognizing which identity to use for a specific sum.
- ✕Forgetting the initial term (r=0) or the final term (r=n) when evaluating sums, especially after differentiation/integration.
📝 Practice Questions
See allQ21.If ∑30r=1 r2(30Cr)230Cr−1
Q20.If the area of the region {(x, y) : −1 ≤x ≤1, 0 ≤y ≤a + e|x| −e−x, a > 0} is e2+8e+1e , then the value of is : (1) 8 (2) 7 (3) 5 (4) 6
Q22.If ∑5r=0 11C22r2r+2 = mn , gcd(m, n) = 1
Q3. Let α, β, γ and δ be the coefficients of x7, x5, x3 and x respectively in the expansion of 5 5 αu + βv = 18 + , x > 1. If u and v satisfy the equations then u + v equals : (x + √x3 −1) (x −√x3 −1) γu + δv = 20 (1) 5 (2) 3 (3) 4 (4) 8
Q24.The sum of all rational terms in the expansion of (1 + 21/2 + 31/2) 6 is equal to
Q6. The product of all the rational roots of the equation (x2 −9x + 11)2 −(x −4)(x −5) = 3, is equal to (1) 14 (2) 21 (3) 28 (4) 7
NCERT Chapters
- Class 11 Maths Ch 8: Binomial Theorem