Sequence and series: sum of $a_1, a_2, a_3, a_4, a_5, a_6, a_7$

Question

I don't have any working of my own to show as I have no idea how to go about doing this question...a hint as to how to start would do.
(I just know that we can take the LCM and make the denominator 7! but I don't know how to proceed further)

Answer 1

HINT: After you put it over a common denominator of $7!$ and multiply through by $7!$ you have

$$a_7+7a_6+42a_5+210a_4+840a_3+2520a_2=3600\;.$$

This can be rewritten

$$a_7+7(a_6+6a_5+30a_4+120a_3+360a_2)=3600\;.$$

Now $3600=7\cdot514+2$, and $0\le a_7<7$, so $a_7$ must be $2$, and

$$a_6+6a_5+30a_4+120a_3+360a_2=514\;.$$

Similarly,

$$514=6a_5+30a_4+120a_3+360a_2=6(a_5+5a_4+20a_3+60a_2)\;;$$

can you continue from there?

Answer 2

Multiplying by 7! and reducing mod 7, we have $$ a_7 \equiv 5 \cdot 6! \equiv -5 \pmod{7} , $$either by Wilson's theorem or by simply doing the computation. Therefore $a_7=2$, so $$ \frac{a_2}{2!} + \dotsb + \frac{a_6}{6!} = \frac{5 \cdot 6! -2 }{7!} = \frac{3598}{7!} = \frac{514}{6!} .$$ Multiplying through by 6! and reducing mod 6, we have $$ a_6 \equiv 514 \equiv 4 \pmod{6}, $$so $a_6 = 4$, and $$ \frac{a_2}{2!} + \dotsb + \frac{a_5}{5!} = \frac{514-4}{6!} = \frac{510}{6!} = \frac{85}{5!} . $$ Proceeding similarly, we have $a_5= 0$ and $$ \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} = \frac{17}{4!}, $$so $a_4 = 1$, and $$ \frac{a_2}{2!} + \frac{a_3}{3!} = \frac{16}{4!} = \frac{4}{3!}, $$ and proceeding as before, we obtain $a_2 = a_3 = 1$. Therefore $$ a_2 + a_3 + a_4 + a_5 + a_6 + a_7 = 1+1+1+0+4+2 = \boxed{9}. $$

Checking, we see that indeed, $$ \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{4}{6!} + \frac{2}{7!} = \frac{5}{7}. $$