- #1

sleepingMantis

- 25

- 5

- Homework Statement
- Prove for any ## 1 \leq m < n## that: $$ \sum_{i = m + 1}^{n} i = \frac{(n - m)(n + m + 1)}{2} $$

- Relevant Equations
- $$ \sum_{1}^{n} i = \frac{n (n + 1)}{2} $$

Hi,

I am self studying induction and came across the following problem. I am stuck on how to proceed (I need to use induction, I know there is a direct proof). My proof attempt is as follows:

Let ## P (m) ## be the proposition that:

$$ \sum_{i = m + 1}^{n} i = \frac{(n - m)(n + m + 1)}{2} $$

Then ## P (1) ## gives:

$$ \sum_{i = 1 + 1}^{n} i = 2 + 3 + ... + n $$

Using the summation equation:

$$ \sum_{i = 2}^{n} i = \frac{n(n + 1)}{2} - 1 $$

$$ \Longleftrightarrow \sum_{i = 2}^{n} i = \frac{n^2 + n - 2}{2} $$

$$ \Longleftrightarrow \sum_{i = 2}^{n} i = \frac{(n - 1)(n + 1 + 1)}{2} $$

Therefore ## P(1) ## is true.

Now assume ## P(k) ## is true such that:

$$ \sum_{i = k + 1}^{n} i = \frac{(n - k)(n + k + 1)}{2} $$

Then ## P(k + 1) ## gives:

$$ \sum_{i = (k + 1) + 1}^{n} i = \sum_{i = (k + 1)}^{n} i - 1 $$

Looking at the right hand side of the equality and using ## P(k) ## :

$$ \frac{(n - k)(n + k + 1)}{2} - 1 $$

This simplifies down to:

$$ \frac{n^2 + n - k^2 - k - 2}{2} $$

My question is how do I proceed from here?

Are there any tricks I should be aware of to convert the numerator into the required form?

Any help is much appreciated!

I am self studying induction and came across the following problem. I am stuck on how to proceed (I need to use induction, I know there is a direct proof). My proof attempt is as follows:

Let ## P (m) ## be the proposition that:

$$ \sum_{i = m + 1}^{n} i = \frac{(n - m)(n + m + 1)}{2} $$

Then ## P (1) ## gives:

$$ \sum_{i = 1 + 1}^{n} i = 2 + 3 + ... + n $$

Using the summation equation:

$$ \sum_{i = 2}^{n} i = \frac{n(n + 1)}{2} - 1 $$

$$ \Longleftrightarrow \sum_{i = 2}^{n} i = \frac{n^2 + n - 2}{2} $$

$$ \Longleftrightarrow \sum_{i = 2}^{n} i = \frac{(n - 1)(n + 1 + 1)}{2} $$

Therefore ## P(1) ## is true.

Now assume ## P(k) ## is true such that:

$$ \sum_{i = k + 1}^{n} i = \frac{(n - k)(n + k + 1)}{2} $$

Then ## P(k + 1) ## gives:

$$ \sum_{i = (k + 1) + 1}^{n} i = \sum_{i = (k + 1)}^{n} i - 1 $$

Looking at the right hand side of the equality and using ## P(k) ## :

$$ \frac{(n - k)(n + k + 1)}{2} - 1 $$

This simplifies down to:

$$ \frac{n^2 + n - k^2 - k - 2}{2} $$

My question is how do I proceed from here?

Are there any tricks I should be aware of to convert the numerator into the required form?

Any help is much appreciated!