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

S = sum = 1 + a + a^2 +...+ a^(n-1)

R = 1 + (1/a) + (1/a)^2 +...+ (1/a)^(n-1)

If we multiply each term in R by a^(n-1), it turns into exactly one of the terms in S.

Therefore,
R*a^(n-1) = S

Taking both sides to the nth power:
R^n * a^[(n-1)n] = S^n

Since: P = 1*a*a^2*..a^(n-1) = a^(n(n-1)/2)
P^2 = a^(n(n-1))

So indeed:
R^n * P^2 = S^n
as desired.

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!!

Answer 3

For the progression a + ar + ar^2 + ... + ar^(n-1) we have
S = a(1 + r + ... + 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 (though admittedly sloppy) algebraic computation completes the proof.