If 1 + r + r^2 + dots + r^n = (1 + r) (1 + r^ | vedclass

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(C) We are given the equation: $1 + r + r^2 + \dots + r^n = (1 + r) (1 + r^2) (1 + r^4) (1 + r^8)$.Using the formula for the sum of a geometric series

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$15$

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(C) We are given the equation: $1 + r + r^2 + \dots + r^n = (1 + r) (1 + r^2) (1 + r^4) (1 + r^8)$.Using the formula for the sum of a geometric series,the left side is $\frac{1 - r^{n+1}}{1 - r}$.So,$\frac{1 - r^{n+1}}{1 - r} = (1 + r) (1 + r^2) (1 + r^4) (1 + r^8)$.Multiplying both sides by $(1 - r)$,we get:$1 - r^{n+1} = (1 - r)(1 + r)(1 + r^2)(1 + r^4)(1 + r^8)$.Using the identity $(1 - x)(1 + x) = 1 - x^2$ repeatedly:$(1 - r)(1 + r) = 1 - r^2$$(1 - r^2)(1 + r^2) = 1 - r^4$$(1 - r^4)(1 + r^4) = 1 - r^8$$(1 - r^8)(1 + r^8) = 1 - r^{16}$.Thus,$1 - r^{n+1} = 1 - r^{16}$.Comparing the exponents,$n + 1 = 16$,which gives $n = 15$.

If $1 + r + r^2 + \dots + r^n = (1 + r) (1 + r^2) (1 + r^4) (1 + r^8)$,then what is the value of $n$?

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