Mathematics: What is the coefficient of xy4 in the expansion of (2x + y)5?

What is the coefficient of xy4 in the expansion

Ответ:

katiekern5207

04.01.2022

Mathematics

The coefficient of xy^4 in the expension of (2x+y)^5 is 10.

Step-by-step explanation:

Given (2x+y)^5 , we have to find the coefficient of xy^4 in the expension of .

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

Here, a = 2x and b = y ,

Substitute above, we get,

$=\sum _{i=0}^5\binom{5}{i}\left(2x\right)^{\left(5-i\right)}y^i$

Using $\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$\quad \:i=0\quad :\quad \frac{5!}{0!\left(5-0\right)!}\left(2x\right)^5y^0$

$\quad \:i=1\quad :\quad \frac{5!}{1!\left(5-1\right)!}\left(2x\right)^4y^1\\\\\quad \:i=2\quad :\quad \frac{5!}{2!\left(5-2\right)!}\left(2x\right)^3y^2\\\\\ \quad \:i=3\quad :\quad \frac{5!}{3!\left(5-3\right)!}\left(2x\right)^2y^3\\\\\quad \:i=4\quad :\quad \frac{5!}{4!\left(5-4\right)!}\left(2x\right)^1y^4\\\\\quad \:i=5\quad :\quad \frac{5!}{5!\left(5-5\right)!}\left(2x\right)^0y^5\\\\$

Adding all terms, we get,

$(2x+y)^5=\quad \frac{5!}{0!\left(5-0\right)!}\left(2x\right)^5y^0+\quad \frac{5!}{1!\left(5-1\right)!}\left(2x\right)^4y^1+\quad \frac{5!}{2!\left(5-2\right)!}\left(2x\right)^3y^2+\quad \frac{5!}{3!\left(5-3\right)!}\left(2x\right)^2y^3+\quad \frac{5!}{4!\left(5-4\right)!}\left(2x\right)^1y^4+\quad \frac{5!}{5!\left(5-5\right)!}\left(2x\right)^0y^5$

On evaluating , we get,

(2x+y)^5=32x^5+80x^4y+80x^3y^2+40x^2y^3+10xy^4+y^5

Thus, the coefficient of xy^4 in the expension of (2x+y)^5 is 10.

0,0(0 оценок)

What is the coefficient of xy4 in the expansion of (2x + y)5?

Ответ:

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question