The second term of the expansion is [tex]-4a^3b[/tex].
Solution:
Given expression:
[tex](a-b)^4[/tex]
To find the second term of the expansion.
[tex](a-b)^4[/tex]
Using Binomial theorem,
[tex](a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}[/tex]
Here, a = a and b = –b
[tex]$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}[/tex]
Substitute i = 0, we get
[tex]$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4[/tex]
Substitute i = 1, we get
[tex]$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b[/tex]
Substitute i = 2, we get
[tex]$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}[/tex]
Substitute i = 3, we get
[tex]$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}[/tex]
Substitute i = 4, we get
[tex]$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}[/tex]
Therefore,
[tex]$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}[/tex]
[tex]=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}[/tex][tex]=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}[/tex]
Hence the second term of the expansion is [tex]-4a^3b[/tex].
