Subjects algebra

Pascal Expansion 2Dc31A

Step-by-step solutions with LaTeX - clean, fast, and student-friendly.

Use the AI math solver

1. **Problem Statement:** Find the value of $k$ in each term of the expansion of $$\left(2x-\frac{1}{3}y\right)^{10}$$ corresponding to the given terms. 2. **Formula:** The general term in the binomial expansion of $$(a+b)^n$$ is given by $$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$ where $r=0,1,2,\ldots,n$. 3. **Apply to our problem:** Here, $a=2x$, $b=-\frac{1}{3}y$, and $n=10$. So, $$T_{r+1} = \binom{10}{r} (2x)^{10-r} \left(-\frac{1}{3}y\right)^r = \binom{10}{r} 2^{10-r} (-1)^r \frac{1}{3^r} x^{10-r} y^r$$ 4. **Match terms:** Each term is of the form $$k x^{10-r} y^r$$ where $$k = \binom{10}{r} 2^{10-r} (-1)^r \frac{1}{3^r}$$ --- **a) Term: $k x^8 y^2$** - Here, $x$ exponent is 8, so $10-r=8 \Rightarrow r=2$. - Calculate $k$: $$k = \binom{10}{2} 2^{8} (-1)^2 \frac{1}{3^2} = 45 \times 256 \times 1 \times \frac{1}{9} = \frac{45 \times 256}{9}$$ - Simplify: $$\frac{45}{9} = 5 \Rightarrow k = 5 \times 256 = 1280$$ **Answer:** $k=1280$ --- **b) Term: $k x^5 y^5$** - $10-r=5 \Rightarrow r=5$ - Calculate $k$: $$k = \binom{10}{5} 2^{5} (-1)^5 \frac{1}{3^5} = 252 \times 32 \times (-1) \times \frac{1}{243} = -\frac{252 \times 32}{243}$$ - Simplify numerator: $$252 \times 32 = 8064$$ - So, $$k = -\frac{8064}{243}$$ **Answer:** $k = -\frac{8064}{243}$ --- **c) Term: $-\frac{5120}{9} x^k y^3$** - $r=3$ (since power of $y$ is 3) - The coefficient from formula: $$k_{coef} = \binom{10}{3} 2^{10-3} (-1)^3 \frac{1}{3^3} = 120 \times 2^{7} \times (-1) \times \frac{1}{27}$$ - Calculate: $$2^{7} = 128$$ $$k_{coef} = 120 \times 128 \times (-1) \times \frac{1}{27} = -\frac{15360}{27} = -\frac{5120}{9}$$ - This matches the given coefficient exactly. - The power of $x$ is $10-r = 10-3=7$. **Answer:** $k=7$ --- **d) Term: $\frac{20}{729} x^2 y^{k+3}$** - Power of $x$ is 2, so $10-r=2 \Rightarrow r=8$ - Power of $y$ is $k+3$, but from formula power of $y$ is $r=8$, so $$k+3=8 \Rightarrow k=5$$ - Check coefficient: $$k_{coef} = \binom{10}{8} 2^{2} (-1)^8 \frac{1}{3^8} = 45 \times 4 \times 1 \times \frac{1}{6561} = \frac{180}{6561} = \frac{20}{729}$$ - Matches given coefficient. **Answer:** $k=5$