Answer:
i need this toooooooooooooooooooo
Step-by-step explanation:
Answer:
I believe the answer is
x + 4 = 6.5
-4 -4
__________
x = 2.5
Step-by-step explanation:
hope this helps!
Let's start from what we know.

Note that:

(sign of last term will be + when n is even and - when n is odd).
Sum is finite so we can split it into two sums, first

with only positive trems (squares of even numbers) and second

with negative (squares of odd numbers). So:

And now the proof.
1) n is even.
In this case, both

and

have

terms. For example if n=8 then:

Generally, there will be:

Now, calculate our sum:



So in this case we prove, that:

2) n is odd.
Here,

has more terms than

. For example if n=7 then:

So there is

terms in

,

terms in

and:

Now, we can calculate our sum:




We consider all possible n so we prove that:
Answer:

Step-by-step explanation:
For this case in order to select the one admiral, captain and commander, all different. We are assuming that the order in the selection no matter, so we can begin selecting an admiral then a captain and then a commander.
So we have 10C1 ways to select one admiral since we want just one
Now we have remaining 9 people and we have 9C1 ways to select a captain since we want a captain different from the admiral selected first
Now we have remaining 8 people and we have 8C1 ways to select a commander since we want a commander different from the captain selected secondly.
The term nCx (combinatory) is defined as:

And by properties 
So then the number of possible way are:

If we select first the captain then the commander and finally the admiral we have tha same way of select 
For all the possible selection orders always we will see that we have 720 to select.