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

Answer 1

#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#
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Answer 2

The given series can be represented using sigma notation as follows:

[ \sum_{n=0}^{6} (-1)^n \times \frac{1}{2^n} ]

This notation indicates the sum of the terms ( (-1)^n \times \frac{1}{2^n} ) from ( n = 0 ) to ( n = 6 ), which covers all terms of the given series up to ( -\frac{1}{128} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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