# Mathematic Induction

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• Jan 20th 2009, 10:25 PM
azuki
Mathematic Induction
Prove by induction that∑_(r=1)^n▒(r+3) 3^r= 3/4[3^n(2n + 5) – 5]

I only manage to get up to

∑_(r=1)^(k+1)▒(r+3) 3^r= 3/4[3^(k+1)(2k + 7) – 5]
• Jan 20th 2009, 10:26 PM
Mush
Quote:

Originally Posted by azuki
Prove by induction that∑_(r=1)^n▒(r+3) 3^r= 3/4[3^n(2n + 5) – 5]

I only manage to get up to

∑_(r=1)^(k+1)▒(r+3) 3^r= 3/4[3^(k+1)(2k + 7) – 5]

What's that funny symbol supposed to be?

Is it this:

$\displaystyle \sum_{r = 1}^{n} (r+3)3^r = \frac{3}{4}[3^n(2n+5)-5]$ ??
• Jan 20th 2009, 10:44 PM
Mush
$\displaystyle \sum_{r = 1}^{n} (r+3)3^r = \frac{3}{4}[3^n(2n+5)-5]$

For $n = 1$

$\displaystyle \sum_{r = 1}^{1} (r+3)3^r = (1+3)3^1 = 4 \times 3 = 12$

And:

$\frac{3}{4}[3^n(2n+5)-5] = \frac{3}{4}[3^1(2(1)+5)-5] =\frac{3}{4}[16] = 12$

Assume it's true for $n = k$

$\displaystyle \sum_{r = 1}^{k} (r+3)3^r = \frac{3}{4}[3^k(2k+5)-5]$

Now for $n = k+1$, we expect that the result is identical to the result for $n = k$ except with one more addition of $(r+3)3^r$ added.

Hence:

$\displaystyle \sum_{r = 1}^{k+1} (r+3)3^r = (k+1+3)3^{k+1}+ \sum_{r = 1}^{k} (r+3)3^r = (k+4)3^k \times 3 + \sum_{r = 1}^{k} (r+3)3^r$

$\displaystyle = (k+4)3^k \times 3 + \frac{3}{4}[3^k(2k+5)-5]$

Now, to complete the proof, you need to manipulate that expression into

$\frac{3}{4}[3^{k+1}(2(k+1)+5)-5]$

Can you do that?
• Jan 20th 2009, 10:52 PM
azuki
Quote:

Originally Posted by Mush
What's that funny symbol supposed to be?

Is it this:

$\displaystyle \sum_{r = 1}^{n} (r+3)3^r = \frac{3}{4}[3^n(2n+5)-5]$ ??

Yes. How do I make the symbol appear correctly in this forum?
• Jan 20th 2009, 10:54 PM
Mush
Quote:

Originally Posted by azuki
Yes. How do I make the symbol appear correctly in this forum?

Look at my previous post. If you click on the equations, you will see the code needed to produce them.
• Jan 20th 2009, 11:00 PM
Mush
Here is the proof, more formally:

$\displaystyle \sum_{r = 1}^{n} (r+3)3^r = \frac{3}{4}[3^n(2n+5)-5]$

For $n = 1$

$\displaystyle \sum_{r = 1}^{1} (r+3)3^r = (1+3)3^1 = 4 \times 3 = 12$

And:

$\frac{3}{4}[3^n(2n+5)-5] = \frac{3}{4}[3^1(2(1)+5)-5] =\frac{3}{4}[16] = 12$

Base case holds.

Assume it's true for $n = k$

$\displaystyle \sum_{r = 1}^{k} (r+3)3^r = \frac{3}{4}[3^k(2k+5)-5]$

Now for $n = k+1$, we expect that the result is identical to the result for $n = k$ except with one more addition of $(r+3)3^r$ added.

Hence:

$\displaystyle \sum_{r = 1}^{k+1} (r+3)3^r = (k+1+3)3^{k+1}+ \sum_{r = 1}^{k} (r+3)3^r = (k+4)3^k \times 3 + \sum_{r = 1}^{k} (r+3)3^r$

$\displaystyle = (k+4)3^{k+1} + \frac{3}{4}[3^k(2k+5)-5]$

$\displaystyle = \frac{3}{4}[3^{k+1}(\frac{2}{3}k+\frac{5}{3})-5]+(k+4)3^{k+1}$

$\displaystyle = \frac{3}{4}[3^{k+1}(\frac{2}{3}k+\frac{5}{3})-5+\frac{4}{3}(k+4)3^{k+1}]$

$\displaystyle = \frac{3}{4}[3^{k+1}(\frac{2}{3}k+\frac{5}{3}+\frac{4}{3}(k+4))-5]$

$\displaystyle = \frac{3}{4}[3^{k+1}(\frac{2}{3}k+\frac{5}{3}+\frac{4k}{3}+\fra c{16}{3})-5]$

$\displaystyle = \frac{3}{4}[3^{k+1}(\frac{6}{3}k+\frac{21}{3})-5]$

$\displaystyle = \frac{3}{4}[3^{k+1}(2k+7)-5]$

$\displaystyle = \frac{3}{4}[3^{k+1}(2(k+1)+5)-5]$

QED.