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]

Printable View

- January 20th 2009, 10:25 PMazukiMathematic 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, 10:26 PMMush
- January 20th 2009, 10:44 PMMush

For

And:

Assume it's true for

Now for , we expect that the result is identical to the result for except with one more addition of added.

Hence:

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

Can you do that? - January 20th 2009, 10:52 PMazuki
- January 20th 2009, 10:54 PMMush
- January 20th 2009, 11:00 PMMush
Here is the proof, more formally:

For

And:

Base case holds.

Assume it's true for

Now for , we expect that the result is identical to the result for except with one more addition of added.

Hence:

QED.