Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P^2 R^n= S^n.?

Answer 1

For the progression a + ar + ar^2 + ... + ar^(n-1) we have
S = a(1 + r + r^2 + ... + r^(n-1)) = a[r^n - 1]/(r - 1),
P = (a^n)*r^(1 + 2 + ... + (n - 1)) = (a^n)*r^[(n-1)*n/2], and
R = 1/a + 1/ar + ... + 1/[ar^(n-1)] =
[r^(n-1) + r^(n-2) +... + 1]/[ar^(n-1)] = (S/a)/[ar^(n-1)] = S/[(a^2)*r^(n-1)].

Now a straight forward (albeit messy) algebraic computation completes your proof.

Answer 2

permit the 1st term of the GP be a and the ratio be r then sum of the words of a GP = a(r^n-a million)/(r-a million) = S made of the words would be a* ar * ar^2.... = a^n * r ^(0+a million+2...n-a million) = a^n * r^[n(n-a million)/2] = P sum of the reciplrocal = [(a million/r)^n - a million]/a(a million/r -a million) = [a million - r^n]/a r^(n-a million) (a million-r) = (r^n-a million)/a r^(n-a million)(r-a million) = R so now we could replace interior the LHS =P^2 R^2 ={a^n * r^[n(n-a million)/2]}^2 {(r^n-a million)/a r^(n-a million)(r-a million)}^2 have been given caught!!