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]
$\displaystyle \displaystyle \sum_{r = 1}^{n} (r+3)3^r = \frac{3}{4}[3^n(2n+5)-5] $
For $\displaystyle n = 1 $
$\displaystyle \displaystyle \sum_{r = 1}^{1} (r+3)3^r = (1+3)3^1 = 4 \times 3 = 12$
And:
$\displaystyle \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 $\displaystyle n = k $
$\displaystyle \displaystyle \sum_{r = 1}^{k} (r+3)3^r = \frac{3}{4}[3^k(2k+5)-5] $
Now for $\displaystyle n = k+1 $, we expect that the result is identical to the result for $\displaystyle n = k$ except with one more addition of $\displaystyle (r+3)3^r $ added.
Hence:
$\displaystyle \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 $
$\displaystyle \displaystyle = (k+4)3^k \times 3 + \frac{3}{4}[3^k(2k+5)-5]$
Now, to complete the proof, you need to manipulate that expression into
$\displaystyle \frac{3}{4}[3^{k+1}(2(k+1)+5)-5] $
Can you do that?
Here is the proof, more formally:
$\displaystyle \displaystyle \sum_{r = 1}^{n} (r+3)3^r = \frac{3}{4}[3^n(2n+5)-5] $
For $\displaystyle n = 1 $
$\displaystyle \displaystyle \sum_{r = 1}^{1} (r+3)3^r = (1+3)3^1 = 4 \times 3 = 12$
And:
$\displaystyle \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 $\displaystyle n = k $
$\displaystyle \displaystyle \sum_{r = 1}^{k} (r+3)3^r = \frac{3}{4}[3^k(2k+5)-5] $
Now for $\displaystyle n = k+1 $, we expect that the result is identical to the result for $\displaystyle n = k$ except with one more addition of $\displaystyle (r+3)3^r $ added.
Hence:
$\displaystyle \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 $
$\displaystyle \displaystyle = (k+4)3^{k+1} + \frac{3}{4}[3^k(2k+5)-5]$
$\displaystyle \displaystyle = \frac{3}{4}[3^{k+1}(\frac{2}{3}k+\frac{5}{3})-5]+(k+4)3^{k+1} $
$\displaystyle \displaystyle = \frac{3}{4}[3^{k+1}(\frac{2}{3}k+\frac{5}{3})-5+\frac{4}{3}(k+4)3^{k+1}] $
$\displaystyle \displaystyle = \frac{3}{4}[3^{k+1}(\frac{2}{3}k+\frac{5}{3}+\frac{4}{3}(k+4))-5] $
$\displaystyle \displaystyle = \frac{3}{4}[3^{k+1}(\frac{2}{3}k+\frac{5}{3}+\frac{4k}{3}+\fra c{16}{3})-5] $
$\displaystyle \displaystyle = \frac{3}{4}[3^{k+1}(\frac{6}{3}k+\frac{21}{3})-5] $
$\displaystyle \displaystyle = \frac{3}{4}[3^{k+1}(2k+7)-5] $
$\displaystyle \displaystyle = \frac{3}{4}[3^{k+1}(2(k+1)+5)-5] $
QED.