# Thread: Product Sums (pi notation)

1. ## Product Sums (pi notation)

Ok, I don't quite have the fonts I need to show this question, so Im going to describe it. Basically, in my math 100 course we're on the summation, products and reimann sums unit. I'm having trouble with the product sums (which look like sigma notation, but with a pi symbol in place of sigma). I'll use 2 number signs instead of the pi thingy because I dont have a font with that symbol.
What is:
98
## 2^i
i=-1

Hopefully that makes sense and hopefully someone can help. Thank you

$\prod\limits_{k = - 1}^{98} {2^k }$

To multiply bases we add exponents.
$2^{\left( {\sum\limits_{k = - 1}^{98} k } \right)}$

3. yes, that is my question. How did you get to that step though? My professor asked us to prove it, but I'm almost totally lost on this subject.

4. Originally Posted by bnay
yes, that is my question. How did you get to that step though? My professor asked us to prove it, but I'm almost totally lost on this subject.
$\prod_{k=-1}^{98}2^k$

Expand it out a little bit...

$\prod_{k=-1}^{98}2^k=2^{-1}\cdot2^0\cdot2^1\cdot2^2\cdot2^3\cdot\dots\cdot2 ^{98}$

We see that each term has the same base, so we can add the exponents and get our solution...

$\therefore\prod_{k=-1}^{98}2^k=2^{-1}\cdot2^0\cdot2^1\cdot2^2\cdot2^3\cdot\dots\cdot2 ^{98}=2^{-1+0+1+\dots+98}$

However, we can express the power in sigma notation:

$2^{-1+0+1+\dots+98}=2^{\left(\sum\limits_{k=-1}^{98}k\right)}$

Thus, we have shown that $\prod_{k=-1}^{98}2^k=2^{\left(\sum\limits_{k=-1}^{98}k\right)}$

$\mathbb{Q.E.D.}$

I hope this makes sense!

--Chris

5. Thank you very much. That was very helpful

6. Hello,

Another way is to play with the logarithm, using this rule : $\ln(ab)=\ln(a)+\ln(b)$

$\log_2 \left(\prod_{k=-1}^{98} 2^k \right)=\sum_{k=-1}^{98} \log_2 (2^k)$

Then use the formula $\log_a (a^b)=b \implies \log_2 (2^k)=k$ :

$\log_2 \left(\prod_{k=-1}^{98} 2^k \right)=\sum_{k=-1}^{98} \log_2 (2^k)=\sum_{k=-1}^{98} k$ (1)

Then, this formula $a^{\log_a(x)}=x$
Here, $x=\prod_{k=-1}^{98} 2^k$

Thus $\prod_{k=-1}^{98} 2^k=2^{\log_2 \left(\prod \limits_{k=-1}^{98} 2^k \right)}=2^{\quad \sum \limits_{k=-1}^{98} k} \quad \text{ by (1)}$

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