Sick-quences (7)

The given series can be rewritten:

$1+(-\frac{1}{11})(-2\times\frac{-1}{2})+\frac{(-\frac{1}{11})^{2}(-2\times\frac{-1}{2}\times-2\times\frac{-3}{2})}{2!}+\frac{(-\frac{1}{11})^{3}(-2\times\frac{-1}{2}\times-2\times\frac{-3}{2}\times-2\times\frac{-5}{2})}{3!}+\;...$

$=1+(\frac{2}{11})(\frac{-1}{2})+\frac{(\frac{2}{11})^{2}(\frac{-1}{2}\times\frac{-3}{2})}{2!}+\;...$

We are now ready to simplify this into the form $(1+x)^{n}$ which equals:

$1+nx+\frac{n(n-1)(x^{2})}{2!}+\frac{n(n-1)(n-2)(x^{3})}{3!}+\;...$ where in this case $n=-\frac{1}{2}$ and $x=\frac{2}{11}$

So the given series can be rewritten: $(1+\frac{2}{11})^{-\frac{1}{2}}$

$=(\frac{13}{11})^{-\frac{1}{2}}=\sqrt\frac{11}{13}$

So $a+b=11+13=\boxed{24}$

I work backwards here but it should still make sense $\begin{aligned} \bigg(\frac{11}{13}\bigg)^{\frac{1}{2}} &= \bigg(\frac{13}{11}\bigg)^{-\frac{1}{2}} \\ &= \bigg(1 +\frac{2}{11}\bigg)^{-\frac{1}{2}} \\ &= 1 + \frac{(-\frac{1}{2})}{1} \times \frac{2}{11} + \frac{(-\frac{1}{2})(-\frac{3}{2})}{1\times 2} \times \bigg(\frac{2}{11}\bigg)^2 + \frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{1\times 2 \times 3} \times \bigg(\frac{2}{11}\bigg)^3 + \dots \\ &= 1 - \left(\dfrac1{11} \right) + \left ( \dfrac1{11} \times \dfrac3{22} \right) - \left( \dfrac1{11} \times \dfrac3{22} \times \dfrac5{33} \right) + \dots \end{aligned}$

The binomial expansion we're allowed to use because $\left| \frac{2}{11} \right| < 1$ . By following the solution the other way, you get the answer from the question.