Question

For some reason would not let me post the summation right

Using Weak Induction


n


E i = (n(n+1))/2


i=1


(a) (1 point) Show that P(1) is true, which completes the base case.


(b) Inductive Step:


i. (1 point) What is your inductive hypothesis?


ii. (1 point) What are you trying to prove?


iii. (2 points) Complete the proof:

Answer 1

Explanation:

We will prove by mathematical induction that, for every natural n,

`\sum^(n)_(i=1)i=(n(n+1))/(2)`

We will prove our base case to be true:

Base case:

For n=1,

`\sum^(n)_(i=1)i=\sum^(1)_(i=1)i=1=(1(1+1))/(2)=(n(n+1))/(2).`

Inductive hypothesis:

Given a natural n,

`\sum^(n)_(i=1)i=(n(n+1))/(2)`

Now, we will assume the inductive hypothesis and then use this assumption, involving n, to prove the statement for n + 1.

Inductive step:

Observe that,

`\sum^(n+1)_(i=1)i=\sum^(n)_(i=1)i+(n+1)=(n(n+1))/(2)+(n+1)=\\\\=(n(n+1)+2(n+1))/(2)=((n+2)(n+1))/(2).`

With this we have proved our statement to be true for n+1.

In conclusion, for every natural n,

`\sum^(n)_(i=1)i=(n(n+1))/(2)`.