Question 1108444
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Note that...
S1 = 1 = 2-1 = 2^1-1
S2 = 3 = 4-1 = 2^2-1
S3 = 7 = 8-1 = 2^3-1
...
Sn = ... = 2^n-1<br>
The "-1" makes little difference when we get up around a million; so essentially we are looking for the largest value of n for which 2^n is less than 1 million.<br>
2^10 = 1024 which is just a bit more than 1000, so 2^20 will be just a bit more than 1 million.<br>
So the largest n for which the series has a sum of less than 1,000,000 is n=19.