giúp mình gấp gấp nha

$a.A=\dfrac{1}{3}-\dfrac{1}{18}-\dfrac{1}{54}-\dfrac{1}{108}-\dfrac{1}{270}-\dfrac{1}{378} \\=\dfrac{1}{3}-\bigg(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+\dfrac{1}{270}+\dfrac{1}{378}\bigg) \\=\dfrac{1}{3}-\bigg(\dfrac{1}{3.6}+\dfrac{1}{6.9}+.....+\dfrac{1}{18.21}\bigg) \\=\dfrac{1}{3}-\dfrac{1}{3}.\bigg(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-....+\dfrac{1}{18}-\dfrac{1}{21}\bigg) \\=\dfrac{1}{3}-\dfrac{1}{3}.\bigg(\dfrac{1}{3}-\dfrac{1}{21}\bigg) \\=\dfrac{1}{3}-\dfrac{1}{3}.\dfrac{6}{21} \\=\dfrac{5}{21} \\b.C=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+.....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}} \\⇒3C=3.\bigg(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+.....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\bigg) \\=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+.....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{99}} \\⇔4C=\bigg(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+.....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\bigg)+\bigg(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+.....+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\bigg) \\=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+.....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}+1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+.....+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}} \\=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+......-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}} \\⇒12C=3.\bigg(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+......-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\bigg) \\=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+.....-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}} \\⇒16C=\bigg(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+......-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\bigg)+\bigg(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+.....-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\bigg) \\=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+......-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}+3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+.....-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}} \\=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}<3 \\⇒16C<3 \\⇒C<\dfrac{3}{16}(đpcm)$