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#