1. Tardigrade
  2. Question
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  4. (1/2 ⋅ 4)+(1 ⋅ 3/2 ⋅ 4 ⋅ 6)+(1 ⋅ 3 ⋅ 5/2 ⋅ 4 ⋅ 6 ⋅ 8)+(1 ⋅ 3 ⋅ 5 ⋅ 7/2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10)+ ldots ldots ldots ldots ldots . . . ∞ is equal to

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Q. $\frac{1}{2 \cdot 4}+\frac{1 \cdot 3}{2 \cdot 4 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 8}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 6 \cdot 8 \cdot 10}+\ldots \ldots \ldots \ldots \ldots . . . \infty \quad$ is equal to

Sequences and Series

Solution:

$ T_n=\frac{1 \cdot 3 \cdot 5 \ldots \ldots \ldots(2 n-1)}{2 \cdot 4 \cdot 6 \ldots \ldots \ldots \ldots .2 n(2 n+2)}[2 n+2)-(2 n+1)$
$T _{ n }=\frac{1.3 .5 \ldots \ldots \ldots(2 n -1)}{2.4 .6 \ldots \ldots \ldots \ldots .2 n }-\frac{1 \cdot 3 \cdot 5 \ldots \ldots \ldots(2 n -1)(2 n +1)}{2.4 .6 \ldots \ldots \ldots \ldots .2 n (2 n +2)} $
$\therefore S _{ n }=\sum T _{ n }=\frac{1}{2}-\frac{1 \cdot 3 \cdot 5 \ldots \ldots \ldots(2 n +1)}{2 \cdot 4 \cdot 6 \ldots \ldots \ldots . .2 n (2 n +2)} \text { Note that } S _{\infty}=\frac{1}{2}$