Question

How do you use the binomial theorem to expand `(2-i)^5`?

Answer 1

`-(38 +41i)`

Explanation:

`(a + b)^n =(nC_0) a^nb^0+ (nC_1) a^(n-1)b^1 +(nC_2) a^(n-2)b^2 +...... (nC_n)a^0b^n`

Here `(a=2 ; b= -i , n= 5)`

We know `nC_r= (n!)/(r!(n-r)!) :. 5C_0=1 , 5C_1=5 , 5C_2=10 ,5C_3=10, 5C_4=5, 5C_5=1`

`(2-i)^5 = 2^5 + 5*2^4*(-i)^1 + 10*2^3*(-i)^2+10*2^2*(-i)^3+5*2^1*(-i)^4+ (-i)^5`
`i^2=-1; i^3=-i,i^4=1,i^5=i`

`32 -80 i - 80 + 40i+10 -i = (32-80+10) + i (-80+40-1) = -38 -41i = -(38+41i)` [Ans]