Sequence

$\begin{aligned}S&=\dfrac5{1\times3\times7}+\dfrac7{3\times5\times9}+\dfrac9{5\times7\times11}+\cdots\\&=\displaystyle\sum_{i=1}^{\infty}\dfrac{2i+3}{(2i-1)(2i+1)(2i+5)}\\&=\displaystyle\sum_{i=1}^{\infty}\dfrac13\left(\dfrac2{(2i-1)(2i+1)}+\dfrac1{(2i+1)(2i+5)}\right)\\&=\dfrac13\left(\displaystyle\sum_{i=1}^{\infty}\dfrac2{(2i-1)(2i+1)}+\displaystyle\sum_{i=1}^{\infty}\dfrac1{(2i+1)(2i+5)}\right)\\&=\dfrac13\left[\displaystyle\lim_{k\to\infty}\sum_{i=1}^k\left(\dfrac1{2i-1}-\dfrac1{2i+1}\right)+\displaystyle\lim_{k\to\infty}\sum_{i=1}^k\dfrac14\left(\dfrac1{2i+1}-\dfrac1{2i+5}\right)\right]\\&=\dfrac13\left[\displaystyle\lim_{k\to\infty}\left(1-\dfrac1{2k+1}\right)+\dfrac14\displaystyle\lim_{k\to\infty}\left(\dfrac13+\dfrac15-\dfrac1{2k+5}\right)\right]\\&=\dfrac13\left[1+\dfrac14\left(\dfrac8{15}\right)\right]\\&=\dfrac13\left(\dfrac{17}{15}\right)\\&=\dfrac{17}{45}\\45S&=17=4^2+1\end{aligned}$

Thus, $n=\boxed4$ .

$\begin{aligned} S & = \frac{5}{1 \times 3 \times 7}+\frac{7}{3 \times 5 \times 9}+\frac{9}{5 \times 7 \times 11}+\cdots \\ & = \sum_{n=0}^\infty \frac{2n+5}{(2n+1)(2n+3)(2n+7)} \quad \quad \small \color{#3D99F6}{\text{By partial fractions}} \\ & = \sum_{n=0}^\infty \left( \frac{1}{3(2n+1)} - \frac{1}{4(2n+3)} - \frac{1}{12(2n+7)}\right) \\ & = \frac{1}{3} \sum_{n=0}^\infty \frac{1}{2n+1} - \frac{1}{4} \left( \sum_{n=0}^\infty \frac{1}{2n+1} - 1 \right) - \frac{1}{12} \left( \sum_{n=0}^\infty \frac{1}{2n+1} - 1 - \frac{1}{3} - \frac{1}{5} \right) \\ & = \left( \frac{1}{3} - \frac{1}{4} - \frac{1}{12} \right) \sum_{n=0}^\infty \frac{1}{2n+1} + \frac{1}{4}(1) + \frac{1}{12}\left(1+\frac{1}{3}+\frac{1}{5} \right) \\ & = \left( 0 \right) \sum_{n=0}^\infty \frac{1}{2n+1} + \frac{1}{4} + \frac{1}{12}\left(\frac{23}{15} \right) \\ & = \frac{17}{45} \end{aligned}$

$\implies 45S = 17 = 16 + 1 = 4^2 + 1 \\ \implies n = \boxed{4}$

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