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How do you find the sum of the finite geometric sequence of #sum_(j=1)^6 32(1/4)^(j-1)# ?

Answer 1

Paisley Johnson

Use a formula found in this reference

#sum_(j=1)^n ar^(j-1)=a(1-r^n)/(1-r); r!=1#

Substituting into the formula, #n=6, a = 32 and r = 1/4#:

#sum_(j=1)^6 32(1/4)^(j-1)=32(1-(1/4)^6)/(1-1/4)#

#sum_(j=1)^6 32(1/4)^(j-1)=32(1-1/4096)/(1-1/4)#

#sum_(j=1)^6 32(1/4)^(j-1)=1365/32#

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Answer 2

Emma Aldridge

To find the sum of the finite geometric sequence given by ( \sum_{j=1}^6 32 \left(\frac{1}{4}\right)^{j-1} ), you can use the formula for the sum of a finite geometric series:

[ S = \frac{a(1 - r^n)}{1 - r} ]

where:

( S ) is the sum of the series,
( a ) is the first term of the sequence,
( r ) is the common ratio,
( n ) is the number of terms in the series.

For the given sequence:

( a = 32 ),
( r = \frac{1}{4} ),
( n = 6 ).

Now, plug these values into the formula and calculate the sum ( S ).

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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.