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Grade 11Algebra

if ai greater than or equal to zero but less than i and a1/1!+a2/2!+a3/3!........+a7/7!=5/7 then find the value of a1+a2+a3+a4........a7

Profile image of Rakesh kumar sharma
10 Years agoGrade 11
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2 Answers

Profile image of Vikas TU
10 Years ago
i > ai > = 0
 
1 > a1 > = 0
2 > a2 > = 0
3 > a3 > = 0
4 > a4 > = 0
5 > a5 > = 0
6 > a6 > = 0
7 > a7 > = 0
 
add all the inequalities we get,
             28 > a1+a2+a3+a4........a7 = > 0
and also given,
Summation (ai/i!) = 5/7 from i = 1 to 7.
Apply A.M >= G.M
on  a1+a2+a3+a4........a7
we get,
 a1+a2+a3+a4........a7 > = 7*(a1.a2.a3......a7)^1/7.............................(1)
 
Apply A.M >= G.M on
a1/1!+a2/2!+a3/3!........+a7/7!
we get,
a1/1!+a2/2!+a3/3!........+a7/7! > = 7*(a1/1!.a2/2!.a3/3!........a7/7!)^1/7
Or
a1/1!+a2/2!+a3/3!........+a7/7! > = (1/(1!*2!*3!.......*7!))*7*(a1.a2.a3......a7)^1/7
Or
7*(a1.a2.a3......a7)^1/7 = (a1/1!+a2/2!+a3/3!........+a7/7!)/(1/(1!*2!*3!.......*7!)
Or
7*(a1.a2.a3......a7)^1/7 = (5/7)*(1!*2!*3!.......*7!)..............(2)
put eqn. (2) in (1).
we get,
 
 a1+a2+a3+a4........a7 =  (5/7)*(1!*2!*3!.......*7!)
                                       = 5*(1!*2!*3!*4!*5!*6!*6!)
is the answer.

Profile image of mycroft holmes
10 Years ago
We have 
 
5/7 = 1/(2!) + 1/3! + 1/(4!)+1/5!+1/6!+5/7!
 
Hence a1+a2+a3+a4+a5+a6+a7 = 10
 
This comes from the fact that every rational between 0 and 1 can be uniquely represented with the numbers (1/k!) with k>1 as bases, just like decimal bases.
 
You can create many such bases e.g. with {1!,2!,3!,....) and the Fibonacci numbers for natural numbers.