Using binomial theorem evaluate each of the following:

Solution:

(i) $(96)^{3}$

$=(100-4)^{3}$

$={ }^{3} C_{0} \times 100^{3} \times 4^{0}-{ }^{3} C_{1} \times 100^{2} \times 4^{1}+{ }^{3} C_{2} \times 100^{1} \times 4^{2}-{ }^{3} C_{3} \times 100^{0} \times 4^{3}$

$=1000000-120000+4800-64$

$=884736$

(ii) $(102)^{5}$

$=(100+2)^{5}$

$={ }^{5} C_{0} \times 100^{5} \times 2^{0}+{ }^{5} C_{1} \times 100^{4} \times 2^{1}+{ }^{5} C_{2} \times 100^{3} \times 2^{2}+{ }^{5} C_{3} \times 100^{2} \times 2^{3}+{ }^{5} C_{4} \times 100^{1} \times 2^{4}+{ }^{5} C_{5} \times 100^{0} \times 2^{5}$

$=10000000000+1000000000+40000000+800000+8000+32$

$=11040808032$

(iii) $(101)^{4}$

$=(100+1)^{4}$

$={ }^{4} C_{0} \times 100^{4}+{ }^{4} C_{1} \times 100^{3}+{ }^{4} C_{2} \times 100^{2}+{ }^{4} C_{3} \times 100^{1}+{ }^{4} C_{4} \times 100^{0}$

$=100000000+4000000+60000+400+1$

$=104060401$

(iv) $(98)^{5}$

$(100-2)^{5}$

$={ }^{5} C_{0} \times 100^{5} \times 2^{0}+-{ }^{5} C_{1} \times 100^{4} \times 2^{1}+{ }^{5} C_{2} \times 100^{3} \times 2^{2}-{ }^{5} C_{3} \times 100^{2} \times 2^{3}+{ }^{5} C_{4} \times 100^{1} \times 2^{4}-{ }^{5} C_{5} \times 100^{0} \times 2^{5}$

$=10000000000-1000000000+40000000-800000+8000-32$

$=9039207968$