Question

Using binomial theorem evaluate each of the following:
(i) (96)³
(ii) (102)⁵
(iii) (101)⁴
(iv) (98)⁵

Answer 1

(i) (96)³
$$\begin{gathered} =(100-4{)}^3 \\ ={}^{3}C_0\times {100}^3\times 4^0-{}^{3}C_1\times {100}^2\times 4^1+{}^{3}C_2\times {100}^1\times 4^2-{}^{3}C_3\times {100}^0\times 4^3 \\ =1000000-120000+4800-64 \\ =884736 \end{gathered}$$

(ii) (102)⁵
$$\begin{gathered} =(100+2{)}^5 \\ ={}^{5}C_0\times {100}^5\times 2^0+{}^{5}C_1\times {100}^4\times 2^1+{}^{5}C_2\times {100}^3\times 2^2+{}^{5}C_3\times {100}^2\times 2^3+{}^{5}C_4\times {100}^1\times 2^4+{}^{5}C_5\times {100}^0\times 2^5 \\ =10000000000+1000000000+40000000+800000+8000+32 \\ =11040808032 \end{gathered}$$

(iii) (101)⁴
$$\begin{gathered} =(100+1{)}^4 \\ ={}^{4}C_0\times {100}^4+{}^{4}C_1\times {100}^3+{}^{4}C_2\times {100}^2+{}^{4}C_3\times {100}^1+{}^{4}C_4\times {100}^0 \\ =100000000+4000000+60000+400+1 \\ =104060401 \end{gathered}$$

(iv) (98)⁵
$$\begin{gathered} (100-2{)}^5 \\ ={}^{5}C_0\times {100}^5\times 2^0+{-}^5{C}_1\times {100}^4\times 2^1+{}^{5}C_2\times {100}^3\times 2^2-{}^{5}C_3\times {100}^2\times 2^3+{}^{5}C_4\times {100}^1\times 2^4-{}^{5}C_5\times {100}^0\times 2^5 \\ =10000000000-1000000000+40000000-800000+8000-32 \\ =9039207968 \end{gathered}$$