Sequence and SeriesMCQMTP Nov 21Question 1798 of 212
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The ratio of sum of first n\displaystyle n natural numbers to sum of cubes of first n\displaystyle n natural numbers is

Options

A3:16\displaystyle 3:16
Bn(n+1)/2\displaystyle n(n+1)/2
C2/n(n+1)\displaystyle 2/n(n+1)
DNone of these
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Correct Answer

Option c2/n(n+1)\displaystyle 2/n(n+1)

All Options:

  • A3:16\displaystyle 3:16
  • Bn(n+1)/2\displaystyle n(n+1)/2
  • C2/n(n+1)\displaystyle 2/n(n+1)
  • DNone of these

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Detailed Solution & Explanation

The sum of the first n\displaystyle n natural numbers is:
S1=n(n+1)2S_1 = \frac{n(n+1)}{2}
The sum of the cubes of the first n\displaystyle n natural numbers is:
S3=[n(n+1)2]2S_3 = \left[\frac{n(n+1)}{2}\right]^2

The ratio is:
Ratio=S1S3=n(n+1)2[n(n+1)2]2=1n(n+1)2=2n(n+1)\text{Ratio} = \frac{S_1}{S_3} = \frac{\frac{n(n+1)}{2}}{\left[\frac{n(n+1)}{2}\right]^2} = \frac{1}{\frac{n(n+1)}{2}} = \frac{2}{n(n+1)}
Hence, **Option C** 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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