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Mathematics

Show that (p+1q)m×(p−1q)n(q+1p)m×(q−1p)n=(pq)m+n\dfrac{\Big(p + \dfrac{1}{q}\Big)^m \times \Big(p - \dfrac{1}{q}\Big)^n}{\Big(q + \dfrac{1}{p}\Big)^m \times \Big(q - \dfrac{1}{p}\Big)^n} = \Big(\dfrac{p}{q}\Big)^{m + n}

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Answer

Given,

⇒(p+1q)m×(p−1q)n(q+1p)m×(q−1p)n=(pq+1q)m×(pq−1q)n(pq+1p)m×(pq−1p)n=(pq+1)mqm×(pq−1)nqn(pq+1)mpm×(pq−1)npn=(pq+1)m×(pq−1)n×pm×pn(pq+1)m×(pq−1)n×qm×qn=pm×pnqm×qn=pm+nqm+n=(pq)m+n.\Rightarrow \dfrac{\Big(p + \dfrac{1}{q}\Big)^m \times \Big(p - \dfrac{1}{q}\Big)^n}{\Big(q + \dfrac{1}{p}\Big)^m \times \Big(q - \dfrac{1}{p}\Big)^n} = \dfrac{\Big(\dfrac{pq + 1}{q}\Big)^m \times \Big(\dfrac{pq - 1}{q}\Big)^n}{\Big(\dfrac{pq + 1}{p}\Big)^m \times \Big(\dfrac{pq - 1}{p}\Big)^n} \\[1em] = \dfrac{\dfrac{(pq + 1)^m}{q^m} \times \dfrac{(pq - 1)^n}{q^n}}{\dfrac{(pq + 1)^m}{p^m} \times \dfrac{(pq - 1)^n}{p^n}} \\[1em] = \dfrac{(pq + 1)^m \times (pq - 1)^n \times p^m \times p^n}{(pq + 1)^m \times (pq - 1)^n \times q^m \times q^n} \\[1em] = \dfrac{p^m \times p^n}{q^m \times q^n} \\[1em] = \dfrac{p^{m + n}}{q^{m + n}} \\[1em] = \Big(\dfrac{p}{q}\Big)^{m + n}.

Hence, proved that (p+1q)m×(p−1q)n(q+1p)m×(q−1p)n=(pq)m+n\dfrac{\Big(p + \dfrac{1}{q}\Big)^m \times \Big(p - \dfrac{1}{q}\Big)^n}{\Big(q + \dfrac{1}{p}\Big)^m \times \Big(q - \dfrac{1}{p}\Big)^n} = \Big(\dfrac{p}{q}\Big)^{m + n}.

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