+23248 a2 = 1 a3 = 1 a4 = 0 a5 = 0 a6 = 34 a7 = 2
5/7 = a2 / 2! + a3 / 3! + a4 / 4! + a5 / 5! + a6 / 6! + a7 / 7!
Start by writing each term with the common denominator 7!;
Left-hand side: (5/7)·[ (6·5·4·3·2·1) / (7·6·5·4·3·2·1) ] = 3600 / 7!
Right-hand side: [ 7·6·5·4·3·a2 + 7·6·5·4·a3 + 7·6·5·a4 + 7·6·a5 + 7·a6 + a7 ] / 7!
Cancel the denominators of 7!: 3600 = 7·6·5·4·3·a2 + 7·6·5·4·a3 + 7·6·5·a4 + 7·6·a5 + 7·a6 + a7
Now, start factoring the right-hand side:
3600 = 7( 6·5·4·3·a2 + 6·5·4·a3 + 6·5·a4 + 6·a5 + a6 ) + a7
3600 = 7( 6( 5·4·3·a2 + 5·4·a3 + 5·a4 + a5 ) + a6 ) + a7
{More later, I have to go to supper ...}
+23248 a2 = 1 a3 = 1 a4 = 0 a5 = 0 a6 = 34 a7 = 2
5/7 = a2 / 2! + a3 / 3! + a4 / 4! + a5 / 5! + a6 / 6! + a7 / 7!
Start by writing each term with the common denominator 7!;
Left-hand side: (5/7)·[ (6·5·4·3·2·1) / (7·6·5·4·3·2·1) ] = 3600 / 7!
Right-hand side: [ 7·6·5·4·3·a2 + 7·6·5·4·a3 + 7·6·5·a4 + 7·6·a5 + 7·a6 + a7 ] / 7!
Cancel the denominators of 7!: 3600 = 7·6·5·4·3·a2 + 7·6·5·4·a3 + 7·6·5·a4 + 7·6·a5 + 7·a6 + a7
Now, start factoring the right-hand side:
3600 = 7( 6·5·4·3·a2 + 6·5·4·a3 + 6·5·a4 + 6·a5 + a6 ) + a7
3600 = 7( 6( 5·4·3·a2 + 5·4·a3 + 5·a4 + a5 ) + a6 ) + a7
{More later, I have to go to supper ...}
+23248 a2 = 1 a3 = 1 a4 = 0 a5 = 0 a6 = 34 a7 = 2
5/7 = a2 / 2! + a3 / 3! + a4 / 4! + a5 / 5! + a6 / 6! + a7 / 7!
Start by writing each term with the common denominator 7!;
Left-hand side: (5/7)·[ (6·5·4·3·2·1) / (7·6·5·4·3·2·1) ] = 3600 / 7!
Right-hand side: [ 7·6·5·4·3·a2 + 7·6·5·4·a3 + 7·6·5·a4 + 7·6·a5 + 7·a6 + a7 ] / 7!
Cancel the denominators of 7!: 3600 = 7·6·5·4·3·a2 + 7·6·5·4·a3 + 7·6·5·a4 + 7·6·a5 + 7·a6 + a7
Now, start factoring the right-hand side:
3600 = 7( 6·5·4·3·a2 + 6·5·4·a3 + 6·5·a4 + 6·a5 + a6 ) + a7
3600 = 7( 6( 5·4·3·a2 + 5·4·a3 + 5·a4 + a5 ) + a6 ) + a7
3600 = 7( 6( 5( 4·3·a2 + 4·a3 + a4 ) + a5 ) + a6 ) + a7
3600 = 7( 6( 5( 4( 3a2 + a3 ) + a4 ) + a5 ) + a6 ) + a7
{More later, I have to go to supper ...}
Continuing:
3600 / 7 = 514 , Remainder = 2
So, 3600 = 514(7) + 2 = 7( 6( 5( 4( 3a2 + a3 ) + a4 ) + a5 ) + a6 ) + a7
Let 2 = a7 and subtract from both sides ---> 514(7) = 7( 6( 5( 4( 3a2 + a3 ) + a4 ) + a5 ) + a6 )
Divide both sides by 7 ---> 514 = 6( 5( 4( 3a2 + a3 ) + a4 ) + a5 ) + a6
514/6 = 85, Remainder = 4
85(6) + 4 = 6( 5( 4( 3a2 + a3 ) + a4 ) + a5 ) + a6
Let 4 = a6 and subtract from both sides ---> 85(6) = 6( 5( 4( 3a2 + a3 ) + a4 ) + a5 )
Divide both sides by 6 ---> 85 = 5( 4( 3a2 + a3 ) + a4 ) + a5
85/5 = 17, Remainder = 0
17(5) + 0 = 5( 4( 3a2 + a3 ) + a4 ) + a5
Let 0 = a5 and subtract from both sides ---> 17(5) = 5( 4( 3a2 + a3 ) + a4 )
Divide both sides by 5 ---> 17 = 4( 3a2 + a3 ) + a4
17/4 = 4, Remainder = 1
4(4) + 1 = 4( 3a2 + a3 ) + a4
Let 1 = a4 and subtract from both sides ---> 4(4) = 4( 3a2 + a3 )
Divide both sides by 4 ---> 4 = 3a2 + a3
This process can be continued one more step, but since a2 and a3 must be whole numbers
---> a3 = 1
---> a2 = 1
In summary: a7 = 2 a6 = 4 a5 = 0 4 = 1 a3 = 1 a2 = 1
Which is a wholy different answer than I had before! But, they both work!
(I have no idea how I got a6 = 34....)
geno3141: But the question says:"There is a single sequence of integers" Your solution looks awfully arbitrary!. If you try the following sequences, you get pretty close to what they want:
0!/2! + 1!/3! + 2!/4! + 3!/5! + 4!/6! + 5!/7! =6/7!!! or even closer:
1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! =181/252, or very nearly 5/7? Or, how about the following concoction!!!!!!!!.
1/2! + 1/3! + 1/4! + 1/5! + 1/6! - 19/7! =5/7 ?????????????
