How do you use sigma notation to write the sum for $1-1/2+1/4-1/8+...-1/128$ ?

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Answer 1 · 2017-03-09T20:46:29

$sum_(k = 0)^7 (-1)^k 1/2^k$

$1-1/2+1/4-1/8+...-1/128$

For the basic pattern, we note this:

$= 1/2^0-1/2^1+1/2^2-1/2^3+...-1/2^k$

In order to work out $k$ , we know that:

$2^k = 128 implies k = 7$

$= 1/2^0-1/2^1+1/2^2-1/2^3+...-1/2^7$

Next, the series is alternating from positive to negative. So we can this:

$= (-1)^0 1/2^0 + (-1)^1 1/2^1+(-1)^2 1/2^2 + (-1)^3 1/2^3+... + (-1)^7 1/(2^7)$

The sum is therefore:

$sum_(k = 0)^7 (-1)^k 1/2^k$

How do you use sigma notation to write the sum for 1-1/2+1/4-1/8+...-1/1281-1/2+1/4-1/8+...-1/128?