Question

show that (p+1/q)^m×(p-1/q)^n upon (q+1/p)^m ×(q-1/p)^n=(p/q)^m+n

Answer 1

Using Left Hand Side

LHS = $\frac{(p + \frac{1}{q}) ^{m}.(p - \frac{1}{q}) ^{n}}{(p + \frac{1}{p}) ^{m}.(p - \frac{1}{p}) ^{n}}$

= $\frac{\frac{(pq + 1) ^{m}}{q^m}.\frac{(pq - 1) ^{n}}{q^n}}{\frac{(pq + 1) ^{m}}{p^m}.\frac{(pq - 1) ^{n}}{p^n}}$

= $\frac{(pq + 1)^m.(pq - 1)^n}{q^m.q^n}$ × $\frac{p^m.p^n}{(pq + 1)^m.(pq - 1)^n}$

= $\frac{p^m.p^n}{q^m.q^n}$

= $\frac{p^{m+n} }{q^{m+n}}$

= $(\frac{p}{q})^{m+n}$

Answer 2

Hence Proved.

Step-by-step explanation:

Given,

$\frac{(p + \frac{1}{q}) ^{m}.(p - \frac{1}{q}) ^{n}}{(p + \frac{1}{p}) ^{m}.(p - \frac{1}{p}) ^{n}} = (\frac{p}{q})^{m+n}$

∴$LHS= \frac{(p + \frac{1}{q}) ^{m}.(p - \frac{1}{q}) ^{n}}{(p + \frac{1}{p}) ^{m}.(p - \frac{1}{p}) ^{n}}$

$= \frac{\frac{(pq + 1) ^{m}}{q^m}.\frac{(pq - 1) ^{n}}{q^n}}{\frac{(pq + 1) ^{m}}{p^m}.\frac{(pq - 1) ^{n}}{p^n}}$

$= \frac{(pq + 1)^m.(pq - 1)^n}{q^m.q^n}\times \frac{p^m.p^n}{(pq + 1)^m.(pq - 1)^n}$

$= \frac{p^m.p^n}{q^m.q^n}$

$= \frac{p^{m+n} }{q^{m+n}}$

$= (\frac{p}{q})^{m+n}$$=RHS$

Hence Proved.