Answer:
40
Step-by-step explanation:
4 times 10 equals 40
i news more sownfjdhe
(-4,-4) B
just for extra space.
Answer:
16:3
Step-by-step explanation:
Thats the way ratios work
We start by proving the statement is true for
. In this case, the left hand side is simply:

The right hand side is:

So the property holds for
.
Now we assume the property is valid for some
and prove that it implies that it is also valid for
. That means we shall obtain:
![1^2 + 4^2 + 7^2 + \dots + (3n-2)^2 + [3(n+1)-2]^2 = \dfrac{(n+1)[6(n+1)^2-3(n+1)-1]}{2}.](https://tex.z-dn.net/?f=1%5E2%20%2B%204%5E2%20%2B%207%5E2%20%2B%20%5Cdots%20%2B%20%283n-2%29%5E2%20%2B%20%5B3%28n%2B1%29-2%5D%5E2%20%3D%20%5Cdfrac%7B%28n%2B1%29%5B6%28n%2B1%29%5E2-3%28n%2B1%29-1%5D%7D%7B2%7D.)
We start by writing the property for
and recognize that we've only added an extra term:
![\underbrace{1^2 + 4^2 + 7^2 + \dots + (3n-2)^2}_{=\dfrac{n(6n^2-3n-1)}{2}} + [3(n+1)-2]^2.](https://tex.z-dn.net/?f=%5Cunderbrace%7B1%5E2%20%2B%204%5E2%20%2B%207%5E2%20%2B%20%5Cdots%20%2B%20%283n-2%29%5E2%7D_%7B%3D%5Cdfrac%7Bn%286n%5E2-3n-1%29%7D%7B2%7D%7D%20%2B%20%5B3%28n%2B1%29-2%5D%5E2.)
We now expand the square and merge everything together in a single fraction:
![\dfrac{n(6n^2-3n-1)}{2} + [3(n+1)-2]^2 = \dfrac{6n^3-3n^2-n}{2} + (3n+1)^2 = \\\\= \dfrac{6n^3-3n^2-n}{2} + (9n^2+6n+1) = \dfrac{6n^3 - 3n^2 -n + 18n^2+12n+2}{2} =\\\\= \dfrac{6n^3+15n^2+11n+2}{2}.](https://tex.z-dn.net/?f=%5Cdfrac%7Bn%286n%5E2-3n-1%29%7D%7B2%7D%20%2B%20%5B3%28n%2B1%29-2%5D%5E2%20%3D%20%5Cdfrac%7B6n%5E3-3n%5E2-n%7D%7B2%7D%20%2B%20%283n%2B1%29%5E2%20%3D%20%5C%5C%5C%5C%3D%20%5Cdfrac%7B6n%5E3-3n%5E2-n%7D%7B2%7D%20%2B%20%289n%5E2%2B6n%2B1%29%20%3D%20%5Cdfrac%7B6n%5E3%20-%203n%5E2%20-n%20%2B%2018n%5E2%2B12n%2B2%7D%7B2%7D%20%3D%5C%5C%5C%5C%3D%20%5Cdfrac%7B6n%5E3%2B15n%5E2%2B11n%2B2%7D%7B2%7D.)
Since the desired expression has a
facto, we may now divide the numerator by
, using, for instance, Ruffini's rule, in order to find it:

So we get:

We now add and subtract
and
in order to obtain the form corresponding to
, since we know that it must appear:

So we finally got:
![1^2 + 4^2 + 7^2 + \dots + (3n-2)^2 + [3(n+1)-2]^2 = \dfrac{(n+1)[6(n+1)^2-3(n+1)-1]}{2},](https://tex.z-dn.net/?f=1%5E2%20%2B%204%5E2%20%2B%207%5E2%20%2B%20%5Cdots%20%2B%20%283n-2%29%5E2%20%2B%20%5B3%28n%2B1%29-2%5D%5E2%20%3D%20%5Cdfrac%7B%28n%2B1%29%5B6%28n%2B1%29%5E2-3%28n%2B1%29-1%5D%7D%7B2%7D%2C)
which is the desired result. So we proved that the statement is true for every integer
.
Answer:
The answer is an (IRA)
Step-by-step explanation:
The IRA means "Individual retirement account"