MCQMediumJEE Main 2026 · 24 January, Shift 2Sum of Series

Mathematics Question from JEE Main 2026 · 24 January, Shift 2

The value of (13+47)+(132+13×47+472)+(133+132×47+13×472+473)+\left(\frac13+\frac47\right) +\left(\frac1{3^2}+\frac13\times\frac47+\frac4{7^2}\right) +\left(\frac1{3^3}+\frac1{3^2}\times\frac47+\frac13\times\frac4{7^2}+\frac4{7^3}\right) +\cdots up to infinite terms is

  • A

    74\dfrac{7}{4}

  • B

    43\dfrac{4}{3}

  • C

    65\dfrac{6}{5}

  • D

    52\dfrac{5}{2}

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Practice more Sum of Series questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Related questions