Question

i, \(I=\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)                                              k, \(K=1+5+5^2+...+5^{100}\)
 

m, \(M=1+3^2+3^4+3^6+...+3^{100}\)

Answer 1

i) \(I=\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}=\dfrac{3^{29}\left(11-3\right)}{4.3^{28}}=\dfrac{3.8}{4}=3.2=6\)

k) \(K=1+5+5^2+...+5^{100}\)

\(\Rightarrow5K=5+5^2+...+5^{101}\)

\(\Rightarrow5K-K=5+5^2+...+5^{101}-1-5-...-5^{100}\)

\(\Rightarrow4K=5^{101}-1\Rightarrow K=\dfrac{5^{101}-1}{4}\)

m) \(M=1+3^2+3^4+...+3^{100}\)

\(9M=3^2+3^4+...+3^{102}\)

\(9M-M=3^2+3^4+...+3^{102}-1-3^2-3^4-...-3^{100}\)

\(8M=3^{102}-1\Rightarrow M=\dfrac{3^{102}-1}{8}\)