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CAT 2023 Slot 3 — QA Question 21

Infinite Geometric ProgressionEasy

The value of 1 + (1+13)14 + (1+13+19)116 + (1+13+19+127)164 + ..., is

Answer & solution

  • A

    15/13

  • B

    27/12

  • 16/11

  • D

    15/8

Solution

Easy

The kk-th bracket is the geometric sum 1+13+⋯+13k−11+\tfrac13+\dots+\tfrac1{3^{k-1}} times 14k\tfrac1{4^k}. Split each bracketed sum across its terms and regroup by powers of 13\tfrac13, collapsing each column as a geometric series.

1

Expand each bracket and group the terms by the power of 13\tfrac13 they carry:

S=1+(14+116+164+⋯ )+13(14+116+164+⋯ )+19(116+164+⋯ )+127(164+⋯ )+⋯\begin{aligned} S=1&+\left(\tfrac14+\tfrac1{16}+\tfrac1{64}+\cdots\right)\\ &+\tfrac13\left(\tfrac14+\tfrac1{16}+\tfrac1{64}+\cdots\right)\\ &+\tfrac19\left(\tfrac1{16}+\tfrac1{64}+\cdots\right)+\tfrac1{27}\left(\tfrac1{64}+\cdots\right)+\cdots \end{aligned}
2

Sum each geometric column (ratio 14\tfrac14):

14+116+⋯=1/41−1/4=13,116+164+⋯=1/161−1/4=112⇒ S=1+13+13⋅13+19⋅112+127⋅148+⋯\begin{aligned} &\tfrac14+\tfrac1{16}+\cdots=\frac{1/4}{1-1/4}=\tfrac13,\quad \tfrac1{16}+\tfrac1{64}+\cdots=\frac{1/16}{1-1/4}=\tfrac1{12}\\ &\Rightarrow\ S=1+\tfrac13+\tfrac13\cdot\tfrac13+\tfrac19\cdot\tfrac1{12}+\tfrac1{27}\cdot\tfrac1{48}+\cdots \end{aligned}
3

Collect the tail from the 19\tfrac19 term onward — it is geometric with ratio 112\tfrac1{12}:

S=1+13+19(1+112+1144+⋯ )=1+13+19⋅11−112=1+13+19⋅1211\begin{aligned} S&=1+\tfrac13+\tfrac19\left(1+\tfrac1{12}+\tfrac1{144}+\cdots\right)\\ &=1+\tfrac13+\tfrac19\cdot\frac{1}{1-\tfrac1{12}}=1+\tfrac13+\tfrac19\cdot\tfrac{12}{11} \end{aligned}
4

Add the fractions:

S=1+13+433=33+11+433=4833=1611S=1+\tfrac13+\tfrac{4}{33}=\frac{33+11+4}{33}=\frac{48}{33}=\frac{16}{11}
S=1611S=\mathbf{\dfrac{16}{11}}
CAT 2023 Slot 3 QA Q21: The value of 1 + 1 + 1 3 1 4 + 1 + 1 3 + 1 9 1 16 + 1 + 1 3 + 1 9 + 1 27 1 64 + ..., is — Solution | TheCATExam