Question

Find the 2th term of the expansion of (a-b)^4.​

Answer 1

The second term of the expansion is
`-4a^3b`.

Solution:

Given expression:

`(a-b)^4`

To find the second term of the expansion.

`(a-b)^4`

Using Binomial theorem,

`(a+b)^(n)=\sum_(i=0)^(n)\left(\begin{array}{l}n \\i\end{array}\right) a^((n-i)) b^(i)`

Here, a = a and b = –b

`$(a-b)^4=\sum_(i=0)^(4)\left(\begin{array}{l}4 \\i\end{array}\right) a^((4-i))(-b)^(i)`

Substitute i = 0, we get

`$(4 !)/(0 !(4-0) !) a^(4)(-b)^(0)=1 \cdot (4 !)/(0 !(4-0) !) a^(4)=a^4`

Substitute i = 1, we get

`$(4 !)/(1 !(4-1) !) a^(3)(-b)^(1)=(4 !)/(3!) a^(3)(-b)=-4 a^(3) b`

Substitute i = 2, we get

`$(4 !)/(2 !(4-2) !) a^(2)(-b)^(2)=(12)/(2 !) a^(2)(-b)^(2)=6 a^(2) b^(2)`

Substitute i = 3, we get

`$(4 !)/(3 !(4-3) !) a^(1)(-b)^(3)=(4)/(1 !) a(-b)^(3)=-4 a b^(3)`

Substitute i = 4, we get

`$(4 !)/(4 !(4-4) !) a^(0)(-b)^(4)=1 \cdot ((-b)^(4))/((4-4) !)=b^(4)`

Therefore,

`$(a-b)^4=\sum_(i=0)^(4)\left(\begin{array}{l}4 \\i\end{array}\right) a^((4-i))(-b)^(i)`

`=(4 !)/(0 !(4-0) !) a^(4)(-b)^(0)+(4 !)/(1 !(4-1) !) a^(3)(-b)^(1)+(4 !)/(2 !(4-2) !) a^(2)(-b)^(2)+(4 !)/(3 !(4-3) !) a^(1)(-b)^(3)+(4 !)/(4 !(4-4) !) a^(0)(-b)^(4)`
`=a^(4)-4 a^(3) b+6 a^(2) b^(2)-4 a b^(3)+b^(4)`

Hence the second term of the expansion is
`-4a^3b`.