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

1 Answer
May 21, 2017

Answer:

#-(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]