# What is the sum of the first 6 terms of this geometric sequence? 1,8,64,512

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Let the term count be

Let the

Then we have:

The rate of increase is far to great for an arithmetic progression so it is a geometric sequence as stated in the question.

Lets have a play!

From our multiplication tables notice that

Lets try

At a guess lets investigate:

So by observation, what is happening is that we have for any term where

Given that

How does

So it works for

Thus the general rule is

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Set

Multiply

Subtract

Factor out

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#"using the sum to n terms for a geometric sequence"#

#•color(white)(x)S_n=(a(1-r^n))/(1-r)=(a(r^n-1))/(r-1)#

#"where a is the first term and r the common ratio"#

#r=a_2/a_1=a_3/a_2= ...... =a_n/a_(n-1)#

#rArrr=8/1=64/8=512/64=8" and "a=1#

#rArrS_6=(1(8^6-1))/7=37449#

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From the reference Geometric Progression we obtain the equation:

We can find the value of a; given that

We can find the value of r; given that

The same reference gives us the equation:

We know that

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