This is a geometric sequence because each term is twice the value of the previous term. So this is what would be called the common ratio, which in this case is 2. Any geometric sequence can be expressed as:
a(n)=ar^(n-1), a(n)=nth value, a=initial value, r=common ratio, n=term number
In this case we have r=2 and a=1 so
a(n)=2^(n-1) so on the sixth week he will run:
a(6)=2^5=32
He will run 32 blocks by the end of the sixth week.
Now if you wanted to know the total amount he runs in the six weeks, you need the sum of the terms and the sum of a geometric sequence is:
s(n)=a(1-r^n)/(1-r) where the variables have the same values so
s(n)=(1-2^n)/(1-2)
s(n)=2^n-1 so
s(6)=2^6-1
s(6)=64-1
s(6)=63 blocks
So he would run a total of 63 blocks in the six weeks.
An extraneous solution is the root of a transformed equation that isn't the root of the original equation.
<h3>What is an extraneous solution?</h3>
Your information is incomplete. Therefore, an overview will be given. It should be noted that an extraneous solution simply means a solution that is gotten from the solving of a problem but isn't a valid solution to the problem.
In order to find whether a solution is extraneous, you'll bed to plug it back into the equation and see if it works.
Learn more about extraneous solution on:
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A translation of 1 unit down
It would be 1/243 for the answer
