Question

Answer 1

3:

\(A=\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{100}}\)

=>\(7A=1+\dfrac{1}{7}+...+\dfrac{1}{7^{99}}\)

=>\(7A-A=1+\dfrac{1}{7}+...+\dfrac{1}{7^{99}}-\dfrac{1}{7}-\dfrac{1}{7^2}-...-\dfrac{1}{7^{100}}\)

=>\(6A=1-\dfrac{1}{7^{100}}=\dfrac{7^{100}-1}{7^{100}}\)

=>\(A=\dfrac{7^{100}-1}{7^{100}\cdot6}\)

2:

\(19M=\dfrac{19^{31}+95}{19^{31}+5}=1+\dfrac{90}{19^{31}+5}\)

\(19N=\dfrac{19^{32}+95}{19^{32}+5}=\dfrac{19^{32}+5+90}{19^{32}+5}=1+\dfrac{90}{19^{32}+5}\)

\(19^{31}+5< 19^{32}+5\)

=>\(\dfrac{90}{19^{31}+5}>\dfrac{90}{19^{32}+5}\)

=>19M>19N

=>M>N