Lovely series

We know,

$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\ldots$

Now,

$nx=\frac{1}{3}$ and $\frac{n(n-1)}{2!}x^2=\frac{5}{36}$

By solving the above equations, we'll get $x=-\frac{1}{2}$ and $n=-\frac{2}{3}$ .

Therefore, $(1+x)^n=\left(\frac{1}{2}\right)^{-\frac{2}{3}}=4^{\frac{1}{3}}$ , so our answer is $\left(4^{\frac{1}{3}}\right)^3 = 4.$

This is just a way to solve the problem without the binomial formula.

We can notice that every single denominator has a multiple of 6 (or a power of 6). If we expand every denominator and group the 6's, we will find a 'tentative' pattern.

$(\dfrac{1}{6^{0} * (1)}$ + $\dfrac{2}{6^{1} * (1)}$ + $\dfrac{10}{6^{2} * (2)}...)$ $^{3}$

This expression is almost the same as the problem, but if we input factorials, then we get an exactly equal expression!

= $(\dfrac{1}{6^{0} * (0!)}$ + $\dfrac{2}{6^{1} * (1!)}$ + $\dfrac{10}{6^{2} * (2!)}...)$ $^{3}$

= $(1+\displaystyle \sum_{n=1}^\infty \dfrac{X}{6^{n}* n!})$ $^{3}$

Note: I left out the 1 for simplicity in future steps, therefore started the infinite sum from 1 instead of 0.

By inspecting, we find that the factors in the numerators are just increasing 3's. Consequently, every factor found in the numerators must be "congruent" in MOD 3 (excluding 1, that's why I left it outside). Therefore, we can use something called triple factorials to rename the series into a simpler expression.

Definition : the triple factorial of n, is the product of positive integers less than n and congruent to n mod 3. So, for example, $8!!! = 8 × 5 × 2$ . By using triple factorials, we only need the largest factor of each numerator.

$2,5,8,11...$

Numerators in the infinite sum are therefore expressed as $(3n-1)!!!$ .

= $(1+\displaystyle \sum_{n=1}^\infty \dfrac{(3n-1)!!!}{6^{n}* n!})$ $^{3}$

Triple factorial are not that famous, but we can transform them to gamma function expressions or simple factorials or Pochhammer notation, which are more popular notations.

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Factorial to Gamma

Where k, x are natural numbers and x < k

$(kn-x)!_{k}$ = $\dfrac{ k^{n} Γ(n+\dfrac{k-x}{k})}{ Γ(\dfrac{k-x}{k})}$

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= $(1+\displaystyle \sum_{n=1}^\infty \dfrac{(3n-1)!!!}{6^{n}* n!})$ $^{3}$

= $(1+\displaystyle \sum_{n=1}^\infty \dfrac{ 3^n * Γ(n+(2/3))}{6^{n}* n! * Γ(2/3)})$ $^{3}$

=( $1 + (2 ^{\dfrac{2}{3}} - 1$ )) $^{3}$

$=4$