Problem 47

Question

Find the term indicated in each expansion. $\left(x^{2}+y\right)^{22} ;$ the term containing $y^{14}$

Step-by-Step Solution

Verified

Answer

The term in the binomial expansion $(x^{2} + y)^{22}$ containing $y^{14}$ is $7315 \cdot x^{16} y^{14}$.

1Step 1: Identify the binomial and the desired term

The binomial is $(x^{2} + y)$ and the power is 22. The term to find is the one containing $y^{14}$. As $y$ only appears once in every term of the expansion, this suggests that $k = 14$.

2Step 2: Find the exponent for $x$

From the binomial theorem, exponent for $a (or x^{2} in this case)$ is $n - k$. For $n = 22$ and $k = 14$, the exponent for $x^{2}$ is $22 - 14 = 8$. Thus, $x^{2}$ is raised to the power of 8, giving $x^{16}$. So the term containing $y^{14}$ will also contain $x^{16}$.

3Step 3: Find the coefficient of the term

The coefficient of the term is given by $\binom{n}{k}$. Substitute $n = 22$ and $k = 14$ to calculate $\binom{22}{14}$, which is 7315.

4Step 4: Write out the term

With the coefficient and the powers of $x$ and $y$, the term containing $y^{14}$ in $(x^{2} + y)^{22}$ is $7315 \cdot x^{16} y^{14}$.

Problem 46

Problem 47