# Math Help - Mathematic Induction

1. ## 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]

2. 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:

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

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

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?

4. 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?

5. 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.

6. 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.