The sum to n terms of 1/1.3+2/1.3.5+3/1.3.5.7+4/1.3.5.7.9+........ is

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MARKS Content Team · Accepted Answer

Correct answer: (C) 1/21-1/(1.3.5........2n+1). t_n=n/(1.3.5. .........2n+1) = 1/2. (2n+1-1)/(1.3.5. .........2n+1) =1/21/(1.3.5......2n-1)-1/(1.3.5.......2n+1) =1/2T_(n-1)- T_n ∴2t_n=T_(n-1)- T_n …1 where T_n=1/(1.3.5.......2n+1) ∴ 2S_n = ∑ n=2 n 2t_n +2t_1 ⇒ 2S_n − 2t_1 = ( T_1 − T_2 )+( T_2 − T_3 )+........+( T_(n−1) −T_n ) ⇒ 2S_n − 2t_1 = T_1 −T_n ⇒ 2S_n = 2t_1 + T_1 −T_n = 2×1/3 +1/(1×3) −1/(1×3×5.....( 2n+1 )) ⇒ S_n = 1/21−1/(1×3×5.......(2n+1))

The sum to $n$ terms of $[\frac{1}{1.3} + \frac{2}{1.3 . 5} + \frac{3}{1.3 . 5.7} + \frac{4}{1.3 . 5.7 . 9} + . . . . . . . .]$ is

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The sum to n terms of 11.3+21.3.5+31.3.5.7+41.3.5.7.9+........ is

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The sum to $n$ terms of $[\frac{1}{1.3} + \frac{2}{1.3 . 5} + \frac{3}{1.3 . 5.7} + \frac{4}{1.3 . 5.7 . 9} + . . . . . . . .]$ is