If S be the sum,P be the product and R the sum of reciprocals of n terms in a G.P,prove that #P^2=(frac{S}R)^n#?

Question

1767 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-16T12:18:23

See below.

The right formulation is

#P^2 = (S/R)^(n+1)#

because

#S = sum_(k=0)^n a^k = (a^(n+1)-1)/(a-1)#

#R = sum_(k=0)^n a^-k = (1/a^(n+1) -1)/(1/a-1) =(a (1-a^(n+1)))/((1-a)a^(n+1))#

then

#S/R = a^n#

and

#P = prod_(k=0)^n a^k = a^(sum_(k=0)^n k) = a^((n(n+1))/2)#

hence

#P^2 = a^(n(n+1)) ne a^(n xx n)#

so the exact formulation is

#P^2 = (S/R)^(n+1)#