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KCET2016MathematicsSequences and Series

The sum of ( 1^ s t n ) terms of the series [ 1² 1 + 1²+2² 1+2 + 1²+2²+3² 1+2+3 + ]

Options

  1. A( n+2 3 )
  2. B( n(n+2) 3 )
  3. C( n(n-2) 3 )
  4. D( n(n-2) 6 )

Correct answer

B. ( n(n+2) 3 )

Step-by-step solution

Given that, [ 1² 1 + 1²+2² 1+2 + 1²+2²+3² 1+2+3 + s ] So, nth term is given by [ array l t_ n = 1²+2²+ +n² 1+2+ +n = n(n+1)(2 n+1) 6 2 n(n+1) = (2 n+1) 3 array ] [ array l Now, t_ n = 1 3 (2 n+ 1) = 1 3 [ 2 n(n+1) 2 +n ]= 1 3 n(n+2) array ]

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