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]
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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: ??
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?
Originally Posted by Mush What's that funny symbol supposed to be? Is it this: ?? Yes. How do I make the symbol appear correctly in this forum?
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.
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.
Last edited by Mush; Jan 21st 2009 at 08:05 AM.
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