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Question Number 167374 by mathlove last updated on 14/Mar/22

	Answered by nurtani last updated on 14/Mar/22

	$P = (7 + 5) (7^{2} + 5^{2}) (7^{4} + 5^{4}) . . . . . (7^{64} + 5^{64})$ $\Leftrightarrow P = \frac{(7 - 5) (7 + 5) (7^{2} + 5^{2}) (7^{4} + 5^{4}) . . . . . (7^{64} + 5^{64})}{(7 - 5)}$ $\Leftrightarrow P = \frac{(7^{2} - 5^{2}) (7^{2} + 5^{2}) (7^{4} + 5^{4}) . . . . . (7^{64} + 5^{64})}{2}$ $\Leftrightarrow P = \frac{(7^{4} - 5^{4}) (7^{4} + 5^{4}) . . . . . (7^{64} + 5^{64})}{2}$ $\Rightarrow P = \frac{(7^{64} - 5^{64}) (7^{64} + 5^{64})}{2} = \frac{7^{128} - 5^{128}}{2}$ $∴ P = (7 + 5) (7^{2} + 5^{2}) (7^{4} + 5^{4}) . . . . . (7^{64} + 5^{64}) = \frac{7^{128} - 5^{128}}{2}$
	Commented by TheSupreme last updated on 15/Mar/22

	$i n g e n e r a l$ $(p + q) \underset{i = 1}{\sum^{N}} (p^{2 i} + q^{2 i}) = \frac{p^{2 N + 1} - q^{2 N + 1}}{p - q}$
	Answered by som(math1967) last updated on 14/Mar/22

	$(7 - 5) P = (7 - 5) (7 + 5) . . . (7^{64} + 5^{64})$ $P = \frac{7^{128} - 5^{128}}{2}$
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