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Little explorations with HP calculators (no Prime)
03-28-2017, 05:46 PM (This post was last modified: 03-28-2017 05:54 PM by Dieter.)
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RE: Little explorations with the HP calculators
(03-21-2017 10:40 PM)pier4r Wrote:  So I got to another problem and I'm stuck.

\[A= \sum_{n=1}^{2014}{n} ~~~\text{and} ~~~ B= \sum_{n=1}^{2014}{n^3}.\]

Find the value of \(\log_{\sqrt A} {B}\)

Now, I know the closed formulas for the two summations, but I want to solve it with the calculator as much as I can.

Better use your brain. ;-)

You know the summation formulas. For any positive upper limit (not just 2014), B is the square of A.
So logAB = 2 and log√AB = 2 · 2 = 4. No calculator required.

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RE: Little explorations with the HP calculators - Dieter - 03-28-2017 05:46 PM

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