Question-185394

Question Number 185394 by mathlove last updated on 21/Jan/23

Answered by mr W last updated on 21/Jan/23

P \times (3^{2^{0}} - 1) = (3^{2^{1}} - 1) (3^{2^{1}} + 1) \dots (3^{2^{n}} + 1)

\dots

P \times (3^{2^{0}} - 1) = (3^{2^{n}} - 1) (3^{2^{n}} + 1)

P \times (3^{2^{0}} - 1) = 3^{2^{n + 1}} - 1

\Rightarrow P = \frac{3^{2^{n + 1}} - 1}{3^{2^{0}} - 1} = \frac{3^{2^{n + 1}} - 1}{2}

Commented by mathlove last updated on 21/Jan/23

t h a n k s