Explain how you can solve 8x8 if you know how to multiply with 4 but not how to multiply with 8
2 answers:
If you only know how to multiply with 4's, then you need to quickly convert the 8's to 4's.
We were given,
So we rewrite each of the 8's as
Perfectly for us,it turns out that the 2's will also give us a 4, so the whole thing becomes,
Now, let us multiply two of the 4's, to obtain,
Finally, let us multiply with the last 4 to get,
We were given,
So we rewrite each of the 8's as
Perfectly for us,it turns out that the 2's will also give us a 4, so the whole thing becomes,
Now, let us multiply two of the 4's, to obtain,
Finally, let us multiply with the last 4 to get,
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Answer:
sum(2^(n+1), for n=1 to 6)
Step-by-step explanation:
To answer this question, you need to know two things:
- what is an expression for the n-th term
- how many terms are there
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The series shown is a geometric series with first term 4 and common ratio 8/4 = 2. The generic form of the n-th term is ...
an = a1×r^(n-1) . . . . first term a1, common ratio r
You can use this form directly in your summation expression, or you can simplify it a bit.
an = 4×2^(n-1) = (2^2)(2^(n-1)) = 2^(n-1+2)
an = 2^(n+1)
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The value 128 is 2^7, so n+1 = 7, or n=6 for that term
Your summation expression could be ...
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<em>Additional comment</em>
The n-th term can also be written as 2×2^n.