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)^nP2=(SR)n?

1 Answer
Cesareo R.
Mar 16, 2018

See below.

Explanation:

The right formulation is

P^2 = (S/R)^(n+1)P2=(SR)n+1

because

S = sum_(k=0)^n a^k = (a^(n+1)-1)/(a-1)S=n∑k=0ak=an+1−1a−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))R=n∑k=0a−k=1an+1−11a−1=a(1−an+1)(1−a)an+1

then

S/R = a^nSR=an

and

P = prod_(k=0)^n a^k = a^(sum_(k=0)^n k) = a^((n(n+1))/2)P=n∏k=0ak=a∑nk=0k=an(n+1)2

hence

P^2 = a^(n(n+1)) ne a^(n xx n)P2=an(n+1)≠an×n

so the exact formulation is

P^2 = (S/R)^(n+1)P2=(SR)n+1

Related questions
Impact of this question
1909 views around the world
You can reuse this answer
Creative Commons License