- #1
Use mathematical induction to prove:
- Thread starter nddewaters
- Start date
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Homework Help Overview
The discussion revolves around using mathematical induction to prove a statement involving summations. Participants are exploring the correct formulation of induction hypotheses and the steps necessary to transition from n = k to n = k + 1.
Discussion Character
- Exploratory, Mathematical reasoning, Assumption checking
Approaches and Questions Raised
- Participants are attempting to clarify their induction hypotheses and the structure of their summations. Some express uncertainty about their current approaches, while others identify potential mistakes in the formulation of their hypotheses.
Discussion Status
There is an ongoing examination of the induction process, with some participants offering corrections and suggestions for improvement. Multiple interpretations of the induction steps are being explored, and guidance has been provided regarding specific errors in the summation notation.
Contextual Notes
Participants have noted issues such as typos in summation indices and the need for clarity in the induction hypothesis. There are also mentions of formatting issues with shared images that may affect readability.
- #2
Mark44
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- 38,106
- 10,661
What have you tried?
- #3
- #4
Mark44
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- 38,106
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You are using induction correctly, but you have a mistake.
Your induction hypothesis for n = k is
\sum_{i = 1}^k \frac{1}{i(i + 1)} = \frac{k}{k + 1}
For n = k + 1 we have
\sum_{i = 1}^{k + 1} \frac{1}{i(i + 1)} = \sum_{i = 1}^k \frac{1}{i(i + 1)} + \frac{1}{(k + 1)(k + 2)}
The last expression on the right is where you made your mistake. You have k + 1. What you should have is the value of 1/(i(i + 1)) when i = k + 1.
BTW, the font size in your attachment is very small, almost too small for my old eyes to read.
Also, if you click either of my equations above, you can see my LaTeX script.
Your induction hypothesis for n = k is
\sum_{i = 1}^k \frac{1}{i(i + 1)} = \frac{k}{k + 1}
For n = k + 1 we have
\sum_{i = 1}^{k + 1} \frac{1}{i(i + 1)} = \sum_{i = 1}^k \frac{1}{i(i + 1)} + \frac{1}{(k + 1)(k + 2)}
The last expression on the right is where you made your mistake. You have k + 1. What you should have is the value of 1/(i(i + 1)) when i = k + 1.
BTW, the font size in your attachment is very small, almost too small for my old eyes to read.
Also, if you click either of my equations above, you can see my LaTeX script.
- #5
- #6
Mark44
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For starters, you have a typo in your summation. It should be
\sum_{i = 1}^n i\cdot i!
The index for your summation is i, not o or 0.
Your induction hypothesis is
\sum_{i = 1}^k i\cdot i! = (k + 1)! - 1
This is the statement when n = k
The statement you're trying to prove, when n = k + 1, is
\sum_{i = 1}^{k + 1} i\cdot i! = (k + 2)! - 1
and not what you have.
\sum_{i = 1}^n i\cdot i!
The index for your summation is i, not o or 0.
Your induction hypothesis is
\sum_{i = 1}^k i\cdot i! = (k + 1)! - 1
This is the statement when n = k
The statement you're trying to prove, when n = k + 1, is
\sum_{i = 1}^{k + 1} i\cdot i! = (k + 2)! - 1
and not what you have.
Last edited:
- #7
nddewaters
- 6
- 0
Mark44 said:
Could you now help me in getting the algebra started that is required to finish the proof?
Thank You
- #8
Mark44
Mentor
- 38,106
- 10,661
\sum_{i = 1}^{k + 1} i\cdot i! = \sum_{i = 1}^{k} i\cdot i! + (k + 1)(k + 1)!
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