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JEE Main202624 January 2026Evening ShiftMathematicsSequences and SeriesActual

( 1 3 + 4 7 )+ ( 1 3² + 1 3 4 7 + 4² 7² )+ ( 1 3³ + 1 3² 4 7 + 1 3 4² 7² + 4³ 7³ )+ upto infinite terms, is equal to

Options

  1. A4 3
  2. B6 5
  3. C5 2
  4. D7 4

Correct answer

C. 5 2

Step-by-step solution

The n -th bracket ( n 1 ) is _ k=0 ^ n ( 1 3 )^ n-k ( 4 7 )^k = (4/7)^ n+1 - (1/3)^ n+1 4/7 - 1/3 = 21 5 [ ( 4 7 )^ n+1 - ( 1 3 )^ n+1 ] . Total sum = 21 5 [ _ n=2 ^ ( 4 7 )^n - _ n=2 ^ ( 1 3 )^n ] = 21 5 [ 16/49 3/7 - 1/9 2/3 ] = 21 5 [ 16 21 - 1 6 ] = 21 5 25 42 = 5 2 .

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