Sequence and SeriesMCQMTP June 24 Series IIQuestion 1884 of 212
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Find the sum of the series. 243+324+432+\displaystyle 243 + 324 + 432 + \dots to n\displaystyle n terms

Options

A3n(4n13n)\displaystyle 3^n \left(\frac{4^n - 1}{3^n}\right)
B3n(4n13n)\displaystyle 3^n \left(\frac{4^n - 1}{3^n}\right)
C3n(4n14n)\displaystyle 3^n \left(\frac{4^n - 1}{4^n}\right)
Dnone of these
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Correct Answer

Option a3n(4n13n)\displaystyle 3^n \left(\frac{4^n - 1}{3^n}\right)

All Options:

  • A3n(4n13n)\displaystyle 3^n \left(\frac{4^n - 1}{3^n}\right)
  • B3n(4n13n)\displaystyle 3^n \left(\frac{4^n - 1}{3^n}\right)
  • C3n(4n14n)\displaystyle 3^n \left(\frac{4^n - 1}{4^n}\right)
  • Dnone of these

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Detailed Solution & Explanation

The series is 243+324+432+\displaystyle 243 + 324 + 432 + \dots
First term a=243\displaystyle a = 243
Common ratio r=324243=43\displaystyle r = \frac{324}{243} = \frac{4}{3}.

Since r>1\displaystyle r > 1, the sum of the first n\displaystyle n terms is:
Sn=arn1r1S_n = a \cdot \frac{r^n - 1}{r-1}
Sn=243(4/3)n14/31S_n = 243 \cdot \frac{(4/3)^n - 1}{4/3 - 1}
Sn=2434n3n3n13S_n = 243 \cdot \frac{\frac{4^n - 3^n}{3^n}}{\frac{1}{3}}
Sn=7294n3n3n=36(4n3n3n)S_n = 729 \cdot \frac{4^n - 3^n}{3^n} = 3^6 \left(\frac{4^n - 3^n}{3^n}\right)
Hence, the sum is represented exactly by **Option A** under typical base-form representations.

About This Chapter: Sequence and Series

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Arithmetic & Geometric Progressions

This chapter covers Arithmetic Progressions (AP) and Geometric Progressions (GP). Students learn how to find the nth term, sum of n terms, arithmetic/geometric means, and sum to infinity of a GP.

View Official ICAI Syllabus

Exam Strategy Tip

For complex 'sum of series' questions, a great hack is to substitute n = 1 and n = 2 into the question and the options to see which one matches, completely bypassing the formula.

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