Expand the given power using the Binomial Theorem. (10k – m)5
Accepted Solution
A:
Answer:[tex](10k - m)^{5}=100000k-50000k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}[/tex]Step-by-step explanation:* Lets explain how to solve the problem- The rule of expand the binomial is:[tex](a+b)^{n}=(a)^{n}+nC1(a)^{n-1}(b)+nC2(a)^{n-2}(b)^{2}+nC3(a)^{n-3}(b)^{3}+...............+(b)^{5}[/tex]∵ The binomial is [tex](10k-m)^{5}[/tex]∴ a = 10k , b = -m and n = 5∴ [tex](10k-m)^{5}=(10k)^{5}+5C1(10k)^{4}(-m)+5C2(10k)^{3}(-m)^{2}+5C3(10k)^{2}(-m)^{3}+5C4(10k)^{1} (-m)^{4}+5C5(10k)^{0}(-m)^{5}[/tex]∵ 5C1 = 5∵ 5C2 = 10∵ 5C3 = 10∵ 5C4 = 5∵ 5C5 = 1∴ [tex](10k-m)^{5}=100000k^{5}+(5)(10000)k^{4}(-m)+(10)(1000)k^{3}(m^{2})+(10)(100)k^{2}(-m^{3})+5(10k)^{1} (m^{4})+(10k)^{0}(-m^{5})[/tex]∴ [tex](10k-m)^{5}=100000k^{5}-50000)k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}[/tex]