$Q.$ If , $$\frac{5}{7}=\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}$$ For , $a_i\in\mathbb Z^+$ & $a_i<i$ . Find $a_2+a_3+......+a_7$ ?
MY APPROACH :
We have , $$\frac{5}{7}=\frac{3!-1}{3!+1}=\frac{3!+1-2}{3!+1}=1-\frac{2}{3!+1} \Rightarrow 1-\frac{2}{7}$$
Now I've : $$1-\frac{2}{7}=\underbrace{\frac{a_2}{2!}+\frac{a_3}{3!}+.........+\frac{a_7}{7!}}_{\Lambda}$$ $$\Lambda=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\left(\frac{1}{4!}-\frac{1}{5!}\right)+\left(\frac{1}{5!}-\frac{1}{6!}\right)+\left(\frac{1}{6!}-\frac{2}{7}\right)$$ OR ,$$\frac{a_2}{2!}+\frac{a_3}{3!}+ \frac{a_4}{4!}+\frac{a_5}{5!}+\frac{a_6}{6!}+\frac{a_7}{7!}=\frac{1}{2!}+\frac{2}{3!}+ \frac{3}{4!}+\frac{4}{5!}+\frac{5}{6!}-\frac{1433}{7!}$$
DOUBT : In my solution , all the terms are following the conditions from the Question except $a_7$ not satisfy $a_7\in\mathbb Z^+$ and $a_7\in\{1,2,3,4,5,6\}$ only by this condition $a_i<i$
$\Rightarrow$ Where am I wrong ?