Mathematic Induction

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January 20th 2009, 09:25 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]
January 20th 2009, 09:26 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:

$\displaystyle \sum_{r = 1}^{n} (r+3)3^r = \frac{3}{4}[3^n(2n+5)-5]$ ??
January 20th 2009, 09:44 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?
January 20th 2009, 09:52 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?
January 20th 2009, 09:54 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.
January 20th 2009, 10:00 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.