use induction to prove sum[i=1,n][5^i+1] = 5^n+2-25/4

[btrfly24]

n
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\
/ 5^i+1 = 5^n+2 - 25/4
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i=1

They want us to use mathematical induction to prove that each statement is true for each positive integer n.

I solved down to this, could someone let me know if I've done it right so far? If so, then I'm stuck from here! Thanks!

5^k+2 - 25/4 + 20^k+2

I can't figure out how to simplify from here. Catherine.

 

[stapel]

Grouping symbols (and web-safe single-line formatting) are generally helpful. What you have posted means:

. . . . .sum[i=1,n] [5ⁱ + 1] = 5ⁿ + 2 - 25/4

Is this what you meant? Or did you mean something more along the lines of the following?

. . . . .sum[i=1,n] [5ⁱ⁺¹] = 5ⁿ⁺² - 25/4

Also, since induction proofs involve three steps (and no "solving"), it would be helpful if you showed all of your work (down to where you say you "solved" something at some point), if we are to be able to check that work.

Thank you.

Eliz.

 

[soroban]

Re: using mathmatical induction

Hello, btrfly24!

If no one is responding, it's because we can't read what you wrote.
. . But I'll take a guess at what you meant . . .

Prove by induction: \(\displaystyle \L\:\sum^n_{i=1}5^{i+1}\;=\;\frac{5^{n+2}\,-\,25}{4}\)

The statement is: \(\displaystyle \L\:5^2\,+\,5^3\,+\,5^4\,+\,\cdots\,+\,5^{n+1}\;=\;\frac{5^{n+2}\,-\,25}{4}\)


\(\displaystyle \text{Verify }S(1):\;5^2\:=\:\frac{5^3\,-\,25}{4}\;\;\Rightarrow\;\;25\,=\,25\) . . . true

\(\displaystyle \text{Assume }S(k):\;5^2\,+\,5^3\,+\,5^4\,+\,\cdots\,+\,5^{k+1}\;=\;\L\frac{5^{k+2}\,-\,25}{4}\)


Add \(\displaystyle 5^{k+2}\) to both sides:

. . \(\displaystyle 5^2\,+\,5^3\,+\,5^4\,+\,\cdots\,+\,5^{k+2}\;=\;\L\frac{5^{k+2}\,-\,25}{4}\)\(\displaystyle \,+\,5^{k+2}\;\) [1]

The right side is: \(\displaystyle \L\:\frac{5^{k+2}\,-\,25\,+\,4\cdot5^{k+2}}{4}\;=\;\frac{5\cdot5^{k+2}\,-\,25}{4}\;=\;\frac{5^{k+3}\,-\,25}{4}\)


Then [1] becomes: \(\displaystyle \:5^2\,+\,5^3\,+\,5^4\,+\,\cdots\,+\,5^{k+2}\;=\;\L\frac{5^{k+3}\,-\,25}{4}\)

This is statement \(\displaystyle S(k+1)\) . . . The induction proof is complete.