Find the Sum upto n terms: 1 3 + 1.5 3.7 + 1.5.9 3.7.11 + 1.5.9.13 3.7.11.15 . . .

Question

Find the Sum upto n terms:      1    3    +      1.5    3.7    +      1.5.9    3.7.11    +      1.5.9.13    3.7.11.15    .  .  .

carponih7 · Accepted Answer

If we call       a    0    =  1      /    3then       a    1    =      a    0    ∗  5      /    7and       a    2    =      a    1    ∗  9      /    11and       a    3    =      a    2    ∗  13      /    15etc.Now, we represent odd numbers as (2n+1). In this case, the numbers follow the pattern of (2n+(2n+1)) for the numerator and the pattern(2n+(2n+3)) for the denominator.So,       a    0    =  1      /    3      a    1    =      a    0    ∗  [  (  4  n  +  1  )      /    (  4  n  +  3  )  ]      a    2    =      a    1    ∗  [  (  4  n  +  1  )      /    (  4  n  +  3  )  ]      a    3    =      a    2    ∗  [  (  4  n  +  1  )      /    (  4  n  +  3  )  ]etc.The nth term (or nth sum) is thus:Let       a    0    =  1      /    3, then      a    n    =  (      a          n      −      1        )  [  (  4  n  +  1  )      /    (  4  n  +  3  )  ] for n&amp;gt;0

Asher Kaufman · Accepted Answer

I assume the sign between terms is a + signLet T = your original question      T          (      n      )        =            n      .1      +                        n          (          n          −          1          )                2            ∗      4              n      .3      +                        n          (          n          −          1          )                2            ∗      4        =            2      n      −      1              2      n      +      1      so let&#039;s see the partern of T      T          (      1      )        =  1      /    3      T          (      2      )        =  3      /    5...      T          (      n      )        =  (  2  n  −  1  )      /    (  2  n  +  1  )  =  1  −  2      /    (  2  n  +  1  )so the sum of T      S          (      n      )        =  1  −      2    3    +  1  −      2    5    +  1  −      2    7    +  .  .  .  +  1  −      2          2      n      +      1        =  n  .1  −  (      2    3    +      2    5    +      2    7    +  .  .  .  +      2          2      n      +      1        )  =  n  −  (      1          1      +              1        2              +      1          2      +              1        2              +      1          3      +              1        2              +  .  .  .  +      1          n      +              1        2              )

Find the Sum upto n terms: 1/3 + 1.5/3.7 + 1.5.9/3.7.11+ (1.5.9.13)/(3.7.11.15) ...

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