Question 588569
The number of chips is in geometric sequence (or geometric progression) where each term is 2 times the previous one (common ratio=2).
The nth question is worth {{{2^n}}} chips.
If the contestant answers the first n questions correctly, the number of accumulated chips will be the sum, S:
{{{S}}}={{{1}}}+{{{2}}}+{{{2^2}}}+{{{2^3}}}+ ..... +{{{2^(n-1)}}}+{{{2^n}}}
Textbooks have formulas to calculate such sums, but I'd rather open my mind than a textbook.
Multiplying times 2:
{{{2S}}}={{{2}}}+{{{2^2}}}+{{{2^3}}}+ ..... +{{{2^(n-1)}}}+{{{2^n}}}+{{{2^(n+1)}}}
and subtracting the expression for S from the expression for 2S:
{{{S}}}={{{2S}}}-{{{S}}}={{{2^(n+1)-1}}}
{{{2^7=128}}} and {{{2^8=256}}}
so the contestant has to aim for n+1=8, and answer the first 7 questions correctly.