If 1+2+2^2+... +2^n=2^{n+1}-1, what is the largest prime fac

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1+2+2^2+... +2^7 = 2^8 – 1 = (2^4+1)(2^4-1) = (2^4+1)(2^2+1)(2^2-1) = (2^4+1)(2^2+1)(2+1)(2-1) = 17*5*3.
17 is the largest prime factor of 1+2+2^2+... +2^7.

Therefore, D is the answer.
Answer: D

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If 1+2+2^2+... +2^n=2^{n+1}-1, what is the largest prime factor of 1+2+2^2+... +2^7?

A. 3
B. 5
C. 13
D. 17
E. 19

There are at least two GMAT-scope ways to find the identity given, therefore it could have been omitted from the question stem!

1st way (the famous trick related to the sum of a finite sequence of terms of a geometric sequence):
$$S = 1 + \underline {{2^1} + {2^2} + \ldots + {2^{n - 1}} + {2^n}} \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,2S = \underline {{2^1} + {2^2} + \ldots + {2^n}} + {2^{n + 1}}\,\,\, = \,\,\,\left( {S - 1} \right) + {2^{n + 1}}$$
$$ \Rightarrow \,\,\,2S - S = {2^{n + 1}} - 1$$


2nd way (based in a not-well-known identity) :
$${x^{n + 1}} - {y^{n + 1}}\,\, = \,\,\,\left( {x - y} \right)\left( {{x^n} + {x^{n - 1}}y + {x^{n - 2}}{y^2} + \ldots + x{y^{n - 1}} + {y^n}} \right)$$
$${2^{n + 1}} - {1^{n + 1}} = \left( {2 - 1} \right)\underbrace {\left( {{2^n} + {2^{n - 1}} + \ldots + 1} \right)}_{ = \,\,S}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,S = {2^{n + 1}} - 1$$

Regards,
Fabio.