Question

fill in the first 7 rows of pascals triangle below: speed hint: use the tab button to move between boxes! use pascals triangle to determine the following: the 2nd term of (a + b)^4 is the 5th term of (a - b)^6 is

Explanation:

Step1: Recall binomial expansion formula

The binomial expansion of $(a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}$, and the coefficients of the expansion are given by the $n$-th row of Pascal's triangle. The $k$-th term (starting from $k = 0$) of $(a + b)^n$ has coefficient $\binom{n}{k}$.

Step2: Find the 2nd term of $(a + b)^4$

For $(a + b)^4$, the rows of Pascal's triangle start counting from $n=0$. The 4 -th row of Pascal's triangle is 1 4 6 4 1. The terms of the binomial expansion of $(a + b)^4$ are $a^{4}+4a^{3}b + 6a^{2}b^{2}+4ab^{3}+b^{4}$. The second term (with $k = 1$) has coefficient 4, so the second term of $(a + b)^4$ is $4a^{3}b$.

Step3: Find the 5th term of $(a - b)^6$

For $(a - b)^6$, the 6 -th row of Pascal's triangle is 1 6 15 20 15 6 1. The binomial expansion of $(a - b)^6=\sum_{k = 0}^{6}\binom{6}{k}a^{6 - k}(-b)^{k}$. The 5th term (with $k = 4$) has coefficient $\binom{6}{4}=\frac{6!}{4!(6 - 4)!}=\frac{6\times5}{2\times 1}=15$, and $(-b)^{4}=b^{4}$, $a^{6-4}=a^{2}$. So the 5th term is $15a^{2}b^{4}$.

Answer:

The 2nd term of $(a + b)^4$ is $4a^{3}b$.
The 5th term of $(a - b)^6$ is $15a^{2}b^{4}$.