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Formulas:
- u_{r+1}=\begin{pmatrix}n \\ r\end{pmatrix}a^{n-r}b^r
- \begin{pmatrix}n \\ r\end{pmatrix}=\frac{n!}{r!(n-r)!}
- What is a binomial?
SolutionA mathematics express that consists of two terms - Expand without using the binomial theorem:
- (x-1)^2
- (x-1)^3
- (x-1)^4
- (2a+1)^4
- Expand (x-1)^4 using the binomial theorem
- Expand (x-1)^7 using the binomial theorem
- Expand (2x-1)^5 using the binomial theorem
- Expand (x-3y)^4 using the binomial theorem
- How many terms when (x+1)^{100} is expanded out?
Solution101 - Evaluate \begin{pmatrix}7 \\ 5\end{pmatrix}
- With a calculator
- Without a calculator
- Simplify:
- \begin{pmatrix}n \\ 0\end{pmatrix}
- \begin{pmatrix}n \\ 1\end{pmatrix}
- \begin{pmatrix}n \\ 2\end{pmatrix}
- Simplify:
- 1^{2n-1}
- (-1)^{3x+1}
- Use Pascal’s triangle to find the coefficients of the terms of (x+1)^6
- The 5th row of Pascal’s triangle is 1, 5, 10, 10, 5, 1. Write down the values of the 6th row of Pascal’s triangle
- Why is it not advisiable to use Pascal’s triangle to determine the coefficients of (2x-y)^5?
- Find the coefficient of a^3b^3 in the expansion of (a+b)^5
- Find the y^4 term in the expansion of (3-y)^7
- In the expansion of ax^3(1+ax)^7, the coefficient of the term in x^4 is 63. Find a
- Find the x^5 term in the expansion of 3x(x+1)^7
- \left(x^2+\frac{3}{x}\right)^7. Find the coefficient of x^8
- In the expansion of (2x-1)^n, the coefficient of the term in x^3 is 16n, where n\in\mathbb{Z}^+. Find n
- The fifth term in the expansion of (x+k)^6 is 3840x^2. Find the possible values of k
- \left(\frac{x^3}{2}+\frac{p}{x}\right)^8. The constant term is 7. Find the possible values of p
- x^2\left(2x^2+\frac{k}{x}\right)^8. The constant term is 7168. Find the possible values of p
- Given \left(1+\frac{1}{3}x\right)^n(3+nx)^2=9+45x+..., find n
- \left(3x^3+\frac{b}{x}\right)^8=6561x^{24}+17496x^{20}+...+kx^0+...
- Find b
- Find k
- \left(1-\frac{3}{2}x\right)^n(2+nx)^2=4-12x-45x^2+216x^3+.... Find n