Mathematics · 2019
JEE Main · 12 April 2019, Shift 1 · Q67
The coefficient of x^18 in the product (1+x)(1-x)^10(1+x+x^2)^9 is:
The coefficient of $\displaystyle x^{18}$ in the product $\displaystyle (1+x)(1-x)^{10}\left(1+x+x^2\right)^9$ is :
Official answer
(1)
$\displaystyle 84$
More from Binomial Theorem
- The term independent of x in the expansion of [(x+1)/(x^2 / 3-x^1 / 3+1)-(x-1)/(x-x^1 / 2)]^10, x ≠ 1, is equal to \ \ \ \.2021
- Suppose Σ_r=0^2023 r^2^2023 C_r=2023 × α × 2^2022. Then the value of α is ____2023
- If Σ_r=0^5 (^11 C_2 r+1)/(2 r+2)=m/n, gcd(m, n)=1, then m-n is equal to ____.2025
- The coefficient of x^18 in the expansion of (x^4-1/x^3)^15 is ____2023
- In the expansion of (1+x)(1-x^2)(1+3/x+3/x^2+1/x^3)^5, x ≠ 0, the sum of the coefficients of x^3 and x^-13 is equal to ____2024
- If the sum of the coefficients of x^7 and x^14 in the expansion of (1/x^3-x^4)^n, x ≠ 0, is zero, then the value of n is ____.2026
- If the coefficient of a^7 b^8 in the expansion of (a+2 b+4 ab)^10 is K · 2^16, then K is equal to \ \ \ \.2021
- Let n ∈ N and [x] denote the greatest integer less than or equal to x. If the sum of (n+1) terms ^n C_0, 3 ·^n C_1, 5 ·^n C_2, 7 ·^n C_3, ….. is…2021
JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.