Q2/122) : find a , b, c,d such that 1/a = 1/b + 1/c + 1/d

We know that

1/n = 1/(n+1) + 1/(n(n+1))  .. (1)

Puting n = a we get

1/a = 1/(a+1) + 1/(a(a+1))

Now putting n = a(a+1)

We get

1/(a(a+1)) = 1/(a(a+1) + 1) + 1/((a(a+1))(a(a+1) + 1) =

= 1/(a^2 + a + 1) + 1/(a(a+1)(a^2+a+1))

Hence b= a+ 1

c= a^2 + a + 1 and d = abc is the solution

for example if a= 2 we have

1/2 = 1/3 + 1/6 ... using 1

1/6 = 1/7 + 1/42 using 1

hence 1/2 = 1/3 + 1/7+ 1/42 ( a= 2 , b= a+1 = 3, c= a^2 + a + 1 = 7 and d= abc = 42)

we can proceed indefinitely to any number of reciprocals. and for any starting reciprocal as well