Question

Suppose \[{a_2},{a_3},{a_4},{a_5},{a_6},{a_7}\] are integers such that,
\[\dfrac{5}{7} = \dfrac{{{a_2}}}{{2!}} + \dfrac{{{a_3}}}{{3!}} + \dfrac{{{a_4}}}{{4!}} + \dfrac{{{a_5}}}{{5!}} + \dfrac{{{a_6}}}{{6!}} + \dfrac{{{a_7}}}{{7!}}\]
Where, \[0 \leqslant a < j\] for \[j = 2,3,4,5,6,7\] . The sum \[{a_2} + {a_3} + {a_4} + {a_5} + {a_6} + {a_7}\] is?

Answer 1

Hint: Here we are given some fractions, which have factorial in its denominator. For the relation we are given above, we have to find the sum of all the unknown variables \[{a_2},{a_3},{a_4},{a_5},{a_6},{a_7}\]. To do so, we see that the denominator values are the product of the previous term denominator and just the next value of the previous value under factorial of the previous term. We use this pattern to solve this question.

Complete step-by-step solution:
We are given that,
\[\dfrac{5}{7} = \dfrac{{{a_2}}}{{2!}} + \dfrac{{{a_3}}}{{3!}} + \dfrac{{{a_4}}}{{4!}} + \dfrac{{{a_5}}}{{5!}} + \dfrac{{{a_6}}}{{6!}} + \dfrac{{{a_7}}}{{7!}}\]
Since we know that \[a! = a \times (a - 1) \times ... \times 1\], we can write the denominators of RHS as,
\[ \Rightarrow \dfrac{5}{7} = \dfrac{{{a_2}}}{{2 \times 3}} + \dfrac{{{a_3}}}{{3 \times 2 \times 1}} + \dfrac{{{a_4}}}{{4 \times 3 \times 2 \times 1}} + \dfrac{{{a_5}}}{{5 \times 4 \times 3 \times 2 \times 1}} + \dfrac{{{a_6}}}{{6 \times 5 \times 4 \times 3 \times 2 \times 1}} + \dfrac{{{a_7}}}{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1!}}\]
We now take common \[\dfrac{1}{2}\] from the RHS, \[\dfrac{1}{3}\]from all terms of RHS except first term and so on,
\[ \Rightarrow \dfrac{5}{7} = \dfrac{1}{2}\left[ {{a_2} + \dfrac{1}{3}\left[ {{a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right]} \right]} \right]\]
We now solve this as,
\[
   \Rightarrow \dfrac{{5 \times 2}}{7} = {a_2} + \dfrac{1}{3}\left[ {{a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right]} \right] \\
   \Rightarrow \dfrac{{10}}{7} = {a_2} + \dfrac{1}{3}\left[ {{a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right]} \right] \\
   \Rightarrow 1 + \dfrac{3}{7} = {a_2} + \dfrac{1}{3}\left[ {{a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right]} \right] \\
 \]
We know that terms after \[{a_2}\] have values less than \[1\], so we get
\[{a_2} = 1\]
Now on solving further,
\[
   \Rightarrow \dfrac{3}{7} = \dfrac{1}{3}\left[ {{a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right]} \right] \\
   \Rightarrow \dfrac{9}{7} = {a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right] \\
   \Rightarrow 1 + \dfrac{2}{7} = {a_3} + \dfrac{1}{4}\left[ {{a_4} + ... + \dfrac{{{a_7}}}{7}} \right] \\
 \]
We can see that terms after \[{a_3}\] have value less than \[1\], so we get
\[{a_3} = 1\]
On solving further we get the values of other unknown integers as,
\[{a_4} = 1\]
\[{a_5} = 0\]
\[{a_6} = 4\]
\[{a_7} = 2\]
Now we are asked to find the value of \[{a_2} + {a_3} + {a_4} + {a_5} + {a_6} + {a_7}\], we find it as,
\[
  {a_2} + {a_3} + {a_4} + {a_5} + {a_6} + {a_7} = 1 + 1 + 1 + 0 + 4 + 2 \\
   \Rightarrow {a_2} + {a_3} + {a_4} + {a_5} + {a_6} + {a_7} = 9 \]
Hence we get the answer as \[9\].

Note: Factorial of a number is a way to write the multiplication of consecutive terms in a decreasing way of that number. It is highly used in Permutation Combination, Binomial theorem, Probability theory and many more areas. It is important to be comfortable with its properties to be able to study math and statistics, especially the topics mentioned above further.