Evaluate the first n terms of 1 ! 1 ( 2 1 ) + 2 ! 1 ⋅ 3 ( 2 1 ) 2 + 3 ! 1 ⋅ 3 ⋅ 5 ( 2 1 ) 3 + ⋯ .

Write t k = ( 1 ⋅ 3 ⋯ ( 2 k − 1 )) / ( k ! 2 k ) and set u k = ( 2 k + 1 ) t k . The recurrence t k + 1 / t k = ( 2 k + 1 ) / [ 2 ( k + 1 )] gives u k − u k − 1 = t k for k ≥ 2 , while u 1 − 1 = t 1 . Telescoping gives S n = u n − 1 = ( 1 ⋅ 3 ⋯ ( 2 n + 1 )) / ( n ! 2 n ) − 1 .

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Evaluate the first $n$ terms of $\frac{1}{1 !} (\frac{1}{2}) + \frac{1 \cdot 3}{2 !} (\frac{1}{2})^{2} + \frac{1 \cdot 3 \cdot 5}{3 !} (\frac{1}{2})^{3} + \dots$ .

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Solution:

Write $t_{k} = (1 \cdot 3 \dots (2 k - 1)) / (k! 2^{k})$ and set $u_{k} = (2 k + 1) t_{k}$ . The recurrence $t_{k + 1} / t_{k} = (2 k + 1) / [2 (k + 1)]$ gives $u_{k} - u_{k - 1} = t_{k}$ for $k \geq 2$ , while $u_{1} - 1 = t_{1}$ . Telescoping gives $S_{n} = u_{n} - 1 = (1 \cdot 3 \dots (2 n + 1)) / (n! 2^{n}) - 1$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: QB

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