The solution of the equation 1 + a + a^2 + a^ | vedclass

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(C) Given the equation: $1 + a + a^2 + \dots + a^x = (1 + a)(1 + a^2)(1 + a^4)$The left side is a geometric series with $x+1$ terms,so its sum is $\fr

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$7$

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(C) Given the equation: $1 + a + a^2 + \dots + a^x = (1 + a)(1 + a^2)(1 + a^4)$The left side is a geometric series with $x+1$ terms,so its sum is $\frac{1 - a^{x+1}}{1 - a}$.Thus,$\frac{1 - a^{x+1}}{1 - a} = (1 + a)(1 + a^2)(1 + a^4)$Multiplying both sides by $(1 - a)$,we get:$1 - a^{x+1} = (1 - a)(1 + a)(1 + a^2)(1 + a^4)$Using the identity $(1 - a)(1 + a) = (1 - a^2)$,we have:$1 - a^{x+1} = (1 - a^2)(1 + a^2)(1 + a^4)$Continuing the pattern $(1 - a^2)(1 + a^2) = (1 - a^4)$:$1 - a^{x+1} = (1 - a^4)(1 + a^4)$Finally,$(1 - a^4)(1 + a^4) = (1 - a^8)$:$1 - a^{x+1} = 1 - a^8$Equating the exponents,$x + 1 = 8$,which gives $x = 7$.

The solution of the equation $1 + a + a^2 + a^3 + \dots + a^x = (1 + a)(1 + a^2)(1 + a^4)$ is given by $x$ is equal to

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