If A(x)*A(y)=A(x+y), calculate A(1)*A(2)*...*A(20)?

Solution:

let x = 1 and y=2

then A(1)*A(2)= A(1+2) = A(3) ..... multiply by A(3)

now A(1)*A(2)*A(3) = A(3)*A(3) = A(6)= A(1+2+3)

==> A(1)*A(2)*A(3) = A(6) ............ multiply by A(4)

==> A(1)*A(2)*A(3)*A(4) = A(6)A(4) = A(10) = A(1+2+3+4)

the A(1)*A(2)......A(20) = A(1+2+3...+20)

Now 1 +2 +3 +...+20 = 20(21)/2 = 210

then A(1)*A(2)*.....A(20) = A(210)

A(x)*A(y) = A(x+y).

To calculate A(1)*A(2)*....A(20)

Solution:

A(x)*A(y) = A(x+y)

A(1)*A(2) = A(1+2) =

A(1)*A(2)*A(3) = A(1+2)*A(3) = A(1+2+3).

A(1)*A(2)*A(3)*A(4) = A(1+2+3+4) .

S0 A(1)*A(2)*A(3)....A(n) = A(1+2+3+4...n) = A(n(n+1)/2),as 1+2+3+4...n = n(n+1)/2.

So A(1)*A(2)*A(3)*....A(20) = A(20*21/2) = A(210)

If the constraint given by the enunciation is A(x)*A(y)=A(x+y), that means that:

A(1)*A(2) = A(1+2)

A(1)*A(2)*A(3)=A(1+2)*A(3)=A(1+2+3)

......................................................

A(1)*A(2)*....*A(20)=A(1+2+3+...20)

But the formula for the sum of the first n terms is:

1+2+.........+n = n*(n+1)/2

So, the sum of the first 20 terms is:

1+2+......20 = (1+20)*20/2 = 21*10 = 210

So, A(1)*A(2)*....*A(20)=A(1+2+3+...20)=A(210).