Let $S_{n} = 1 \cdot (n-1) + 2 \cdot (n-2) + 3 \cdot (n-3) + \dots + (n-1) \cdot 1$,for $n \geq 4$. The sum $\sum_{n=4}^{\infty} \left( \frac{2 S_{n}}{n!} - \frac{1}{(n-2)!} \right)$ is equal to:

If $a = \sum\limits_{n = 0}^\infty {\frac{{{x^{3n}}}}{{(3n)!}}} ,\,b = \sum\limits_{n = 1}^\infty {\frac{{{x^{3n - 2}}}}{{(3n - 2)!}}} $ and $c = \sum\limits_{n = 1}^\infty {\frac{{{x^{3n - 1}}}}{{(3n - 1)!}}} $ then the value of ${a^3} + {b^3} + {c^3} - 3abc = $

If $\cosh (x-\log 3)=\sinh x$,then $x=$

The sum of the series $C = 1 + \frac{\cos x}{1!} + \frac{\cos 2x}{2!} + \frac{\cos 3x}{3!} + \dots$ and $S = \frac{\sin x}{1!} + \frac{\sin 2x}{2!} + \frac{\sin 3x}{3!} + \dots$ is equal to

Find the sum to infinity of the series $\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$

The money invested in a company is compounded continuously. ₹ $400$ invested today becomes ₹ $800$ in $6$ years. What will it become at the end of $33$ years? (Given $\sqrt{2} \approx 1.4142$)