Sequence and SeriesMCQMTP June 24 Series IIQuestion 1884 of 213
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Find the sum of the series. $243 + 324 + 432 + \dotsto\displaystyle ton$ terms

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option a — 3n(4n−13n)\displaystyle 3^n \left(\frac{4^n - 1}{3^n}\right)

All Options:

  • A3n(4n−13n)\displaystyle 3^n \left(\frac{4^n - 1}{3^n}\right)
  • B3n(4n−13n)\displaystyle 3^n \left(\frac{4^n - 1}{3^n}\right)
  • C3n(4n−14n)\displaystyle 3^n \left(\frac{4^n - 1}{4^n}\right)
  • Dnone of these

Detailed Solution & Explanation

The series is $243 + 324 + 432 + \dots<br>Firstterm\displaystyle <br>First terma = 243<br>Commonratio\displaystyle <br>Common ratior = 324243=43<span\displaystyle \frac{324}{243} = \frac{4}{3}<span class="katex">.<br><br>Since\displaystyle .<br><br>Sincer > 1$, the sum of the first $n$ terms is:
Sn=a⋅rn−1r−1S_n = a ⋅\displaystyle \cdot rn−1r−1\displaystyle \frac{r^n - 1}{r-1}
Sn=243⋅(4/3)n−14/3−1S_n = 243 ⋅\displaystyle \cdot (4/3)n−14/3−1\displaystyle \frac{(4/3)^n - 1}{4/3 - 1}
Sn=243⋅4n−3n3n13S_n = 243 ⋅\displaystyle \cdot 4n−3n3n13\displaystyle \frac{\frac{4^n - 3^n}{3^n}}{\frac{1}{3}}
Sn=729⋅4n−3n3n=36(4n−3n3n)S_n = 729 ⋅\displaystyle \cdot 4n−3n3n=36\displaystyle \frac{4^n - 3^n}{3^n} = 3^6 \left(4n−3n3n\displaystyle \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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