Questions

The sum to $n$ terms of the series
$\frac{1}{1^{3}} + \frac{1 + 2}{1^{3} + 2^{3}} + \frac{1 + 2 + 3}{1^{3} + 2^{3} + 3^{3}} + \dots$ is

Remember concepts with our Masterclasses.

80k Users

60 mins Expert Faculty Ask Questions

nn+1

n2(n+1)

2nn+1

nn(n+1)

Ready to Test Your Skills?

Check Your Performance Today with our Free Mock Tests used by Toppers!

detailed solution

Correct option is C

If tn denotes the nth term of the series, thentn=1+2+3+⋯+n13+23+33+⋯+n3=12n(n+1)14n2(n+1)2=2n(n+1)=21n−1n+1⇒ ∑k=1n tk=2112+1213+⋯+ 1n−1n+1 =21−1n+1=2nn+1

free study material

JEE

JEE Mock Tests

JEE study guide

JEE Revision Notes

JEE Important Questions

JEE Sample Papers

JEE Previous Year's Papers

NEET

NEET previous year’s papers

NEET important questions

NEET sample papers

NEET revision notes

NEET study guide

NEET mock tests

CUET

CUET previous year’s papers

CUET important questions

CUET sample papers

CUET revision notes

CUET study guide

CUET mock tests

CBSE

CBSE previous year’s papers

CBSE important questions

CBSE sample papers

CBSE revision notes

CBSE study guide

CBSE mock tests

NCERT solutions

Class 12 NCERT Solutions Class 11 NCERT Solutions Class 10 NCERT Solutions Class 9 NCERT Solutions Class 8 NCERT Solutions Class 7 NCERT Solutions Class 6 NCERT Solutions

NCERT exempler

Class 12 NCERT Exemplar Class 11 NCERT Exemplar Class 10 NCERT Exemplar Class 9 NCERT Exemplar Class 8 NCERT Exemplar Class 7 NCERT Exemplar Class 6 NCERT Exemplar

State Board

Rajasthan State Board West Bengal State Board Gujarat State Board Telengana State Board Andhra Pradesh State Board Uttar Pradesh State Board

Subject

Maths Science Physics Chemistry Economics Accounting Business Studies

2023 | www.infinitylearn.com

The sum to n terms of the series113+1+213+23+1+2+313+23+33+…is

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

The sum to $n$ terms of the series
$\frac{1}{1^{3}} + \frac{1 + 2}{1^{3} + 2^{3}} + \frac{1 + 2 + 3}{1^{3} + 2^{3} + 3^{3}} + \dots$ is