Solveeit Logo
Login

Question

Mathematics Question on Sequences and Series

Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn.

Answer

Let the G.P. be a, ar, ar2 , ar3 , … arn-1…
According to the given information,
S=a(rn−1)r−1S =\frac{ a(r^n - 1) }{ r - 1}
P=an×r1+2+...+n−1P = a^n × r^{1 + 2 + ... + n - 1}
=anrn(n−1)2= a^n r^{\frac{n(n - 1) }{ 2} } [ ∵ Sum of first n natural numbers is n (n+1)2\frac{(n + 1) }{ 2} ]
R=1a+1ar+...+1arn−1R = \frac{1 }{ a} + \frac{1 }{ ar} + ... + \frac{1 }{ ar ^{n - 1}}
=rn−1+rn−2+....r+1arn−1= \frac{r^{ n - 1} + r ^{n - 2} + .... r + 1}{ar ^{n - 1}}
=1(rn−1)(r−1)×1arn−1=\frac{ 1(r^ n - 1) }{ (r - 1) }× \frac{1 }{ ar ^n - 1 } [ ∵ 1, r, ...r n - 1 forms a G.P.]
=rn−1arn−1(r−1)=\frac{r^{ n - 1 }}{ ar ^{n - 1}(r - 1)}
∴ P2Rn=a2nrn(n−1)(rn−1)nanrn(n−1)(r−1)n∴ \,P^2R^n = a^{2n} r^{n(n - 1)}\frac{(r^n-1)^n }{a^nr^{n(n-1)}(r - 1)^n}
=an(rn−1)n(r−1)n= \frac{a^n(r^n - 1)^n }{(r - 1)^n}
=[a(rn−1)(r−1)]n= \frac{[a(r^n - 1) }{ (r - 1)]}^n
=Sn= S^n
Hence, P2 Rn = Sn.