$\frac{a_{1}}{1 + a_{1}} + \frac{a_{2}}{\left( 1 + a_{1} \right)\left( 1 + a_{2} \right)} + \frac{a_

Question

$\frac{a_{1}}{1 + a_{1}} + \frac{a_{2}}{\left( 1 + a_{1} \right)\left( 1 + a_{2} \right)} + \frac{a_{3}}{\left( 1 + a_{1} \right)\left( 1 + a_{2} \right)\left( 1 + a_{3} \right)} + \ldots. + \frac{a_{n}}{\left( 1 + a_{1} \right)\left( 1 + a_{2} \right)\ldots\left( 1 + a_{n} \right)}$ is equal to

Accepted Answer

Sol. [Key Concept: Telescoping: $\frac{a_{k}}{(1 + a_{1}\cdots\ a_{k - 1})(1 + a_{1}\cdots\ a_{k})}$ is a difference of consecutive terms.]$$\left( 1 - \frac{1}{1 + a_{1}} ight) + \left( \frac{1}{1 + a_{1}} - \frac{1}{\left( 1 + a_{1} ight)\left( 1 + a_{2} ight)} ight)$$$$\begin{matrix} & \ + \left( \frac{1}{\left( 1 + a_{1} ight)\left( 1 + a_{2} ight)} - \frac{1}{\left( 1 + a_{1} ight)\left( 1 + a_{2} ight)\left( 1 + a_{3} ight)} ight) + \ldots + \ & \left( \frac{1}{\left( 1 + a_{1} ight)\left( 1 + a_{2} ight)\ldots\left( 1 + a_{n - 1} ight)} - \frac{1}{\left( 1 + a_{1} ight)\left( 1 + a_{2} ight)\ldots\left( 1 + a_{n} ight)} ight) \ = & 1 - \frac{1}{\left( 1 + a_{1} ight)\left( 1 + a_{2} ight)\left( 1 + a_{3} ight)\ldots\left( 1 + a_{n} ight)} \end{matrix}$$

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