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When

$\sum^{60}_{r=0} 2^r = 2^n-1$

what is n?

I forgot to mention that I can't use a calculator on this

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    Any thoughts? Hint: try it with numbers smaller than $60$. – 2017-02-24
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    try with low numbers. For example, change the 60 to 1,2,3,4.. – 2017-02-24
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    It appears to me that the person who gave you the exercise expects you to have studied the formula for the geometric series; so, I suggest you go look it up either on wikipedia or in the schoolbook. – 2017-02-24

3 Answers 3

3

Hint: $$\sum_{r=0}^n2^r=2^{n+1}-1$$ use this formula

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    thank you for the $-1$!!!! – 2017-02-24
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    your formula is wrong above the Exponent must be $$n+1$$ – 2017-02-24
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    I gave you a +1. This is the answer I wan't since I can't use a calculator. – 2017-02-24
2

Hint:

$$2^r=(2-1)2^r=2^{r+1}-2^r$$

So then what we have is,

$$(2^1-2^0)+(2^2-2^1)+(2^3-2^2)+\cdots (2^{61}-2^{60})$$

2

Hint:

$$\begin{align}2^0&=2^1-1\\2^0+2^1&=2^2-1\\2^0+2^1+2^2&=2^3-1\\2^0+2^1+2^2+2^3&=2^4-1\end{align}$$

Can you prove this pattern continues to hold true?