Question

{"given":"The series is grouped so that the $n$-th group ($n = 1, 2, 3, \\ldots$) is $$G_n = \\sum_{k=0}^{n} \\frac{1}{3^{n-k}}\\cdot\\frac{4^k}{7^k} = \\sum_{k=0}^{n} b^{n-k}\\,a^k, \\quad a = \\frac{4}{7},\\ b = \\frac{1}{3}.$$","key_observation":"Using the identity $\\sum_{k=0}^{n} a^k b^{n-k} = \\dfrac{a^{n+1}-b^{n+1}}{a-b}$ (for $a \\neq b$), each group simplifies to a ratio of geometric terms, and the total sum becomes a difference of two convergent geometric series.","option_analysis":[{"label":"(A)","text":"$\\dfrac{5}{2}$","verdict":"correct","explanation":"With $a = \\frac{4}{7}$, $b = \\frac{1}{3}$, we get $a - b = \\frac{5}{21}$. The total sum is $S = \\frac{1}{a-b}\\left(\\frac{a^2}{1-a} - \\frac{b^2}{1-b}\\right) = \\frac{21}{5}\\left(\\frac{16}{21} - \\frac{1}{6}\\right) = \\frac{21}{5}\\cdot\\frac{25}{42} = \\frac{5}{2}$."},{"label":"(B)","text":"$\\dfrac{7}{4}$","verdict":"incorrect","explanation":"This value does not result from the correct evaluation of either geometric sum; it would require $\\frac{a^2}{1-a} - \\frac{b^2}{1-b}$ to equal a different value."},{"label":"(C)","text":"$\\dfrac{4}{3}$","verdict":"incorrect","explanation":"This is smaller than the correct answer and does not match the sum of the two infinite geometric series involved."},{"label":"(D)","text":"$\\dfrac{6}{5}$","verdict":"incorrect","explanation":"This value is also inconsistent with the computed result of $\\frac{5}{2}$."}],"answer":"(A)","formula_steps":["Let $a = \\dfrac{4}{7}$ and $b = \\dfrac{1}{3}$, so $a - b = \\dfrac{4}{7} - \\dfrac{1}{3} = \\dfrac{5}{21}$.","The $n$-th group is $G_n = \\dfrac{a^{n+1} - b^{n+1}}{a - b}$, so $S = \\displaystyle\\sum_{n=1}^{\\infty} \\frac{a^{n+1} - b^{n+1}}{a-b} = \\frac{1}{a-b}\\left(\\frac{a^2}{1-a} - \\frac{b^2}{1-b}\\right)$.","$\\dfrac{a^2}{1-a} = \\dfrac{16/49}{3/7} = \\dfrac{16}{49}\\cdot\\dfrac{7}{3} = \\dfrac{16}{21}$","$\\dfrac{b^2}{1-b} = \\dfrac{1/9}{2/3} = \\dfrac{1}{9}\\cdot\\dfrac{3}{2} = \\dfrac{1}{6}$","$S = \\dfrac{21}{5}\\left(\\dfrac{16}{21} - \\dfrac{1}{6}\\right) = \\dfrac{21}{5}\\cdot\\dfrac{32-7}{42} = \\dfrac{21}{5}\\cdot\\dfrac{25}{42} = \\dfrac{5}{2}$"]}