Use mathematical induction to prove:

[nddewaters]

Please help! Use mathematical induction to prove.

 

[Mark44]

What have you tried?

 

[nddewaters]

What have you tried?

This is as far as I can go. Not totally sure if the Induction Hypothesis is used the right way. Any help is appreciated...

ND

 

[Mark44]

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.

 

[nddewaters]

Please help! Use mathematical induction to prove.

For the second question I have ried the following... please correct me if I'm wrong or some help as to how to proceed further. Thank You

PS- I am sorry for the small images..

 

[Mark44]

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.

 

[nddewaters]

For starters, you have a typo in your summation. It should be
$$\sum_{o = 1}^n i\cdot i!$$

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.

Could you now help me in getting the algebra started that is required to finish the proof?
Thank You

 

[Mark44]

$$\sum_{i = 1}^{k + 1} i\cdot i! = \sum_{i = 1}^{k} i\cdot i! + (k + 1)(k + 1)!$$