Sequence and SeriesMCQMTP June 24 Series IIQuestion 1883 of 213
All Questions

The sum of the following series 4+44+444+…\displaystyle 4 + 44 + 444 + \dots to n\displaystyle n terms is:

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

✅ Option a — 49[10(10n−1)9−n]\displaystyle \frac{4}{9}\left[\frac{10(10^n - 1)}{9} - n\right]

All Options:

  • A49[10(10n−1)9−n]\displaystyle \frac{4}{9}\left[\frac{10(10^n - 1)}{9} - n\right]
  • B49[10(10n+1−1)9+n]\displaystyle \frac{4}{9}\left[\frac{10(10^{n+1} - 1)}{9} + n\right]
  • C10(10n−1)9+n\displaystyle \frac{10(10^n - 1)}{9} + n
  • DNone of these

Detailed Solution & Explanation

Let the sum be Sn=4+44+444+…\displaystyle S_n = 4 + 44 + 444 + \dots
Sn=4[1+11+111+… ]S_n = 4 [1 + 11 + 111 + \dots]
Sn=49[9+99+999+… ]S_n = 49\displaystyle \frac{4}{9} [9 + 99 + 999 + \dots]
Sn=49[(10−1)+(102−1)+… ]S_n = 49\displaystyle \frac{4}{9} [(10-1) + (10^2-1) + \dots]
Sn=49[(10+102+⋯+10n)−n]S_n = 49\displaystyle \frac{4}{9} [(10 + 10^2 + \dots + 10^n) - n]
Sn=49[10(10n−1)9−n]S_n = 49\displaystyle \frac{4}{9} \left[10(10n−1)9−n\displaystyle \frac{10(10^n-1)}{9} - n\right]
Hence, **Option A** is the correct answer.

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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