The Sequence:
1 , 1+2 , 1+2+3 , 1+2+3+4, ...
here, every term is an AP. finding the general formula for the sum of the elements of each term is:
Sₙ = 
<em> [where x = number of term, a = first term and d = common difference]</em>
here, the first term is always 1 and so is the common difference.
Sₙ = 
Sₙ = 
= 
which is the formula for a general term in our series
now, we need to find the sum of the first n terms of this series
![\displaystyle\sum_{x=1}^{n} [\frac{1}{2}(x + x^{2})]](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bx%3D1%7D%5E%7Bn%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28x%20%2B%20x%5E%7B2%7D%29%5D)
![\displaystyle\frac{1}{2} [\sum_{x=1}^{n} (x) + \sum_{x=1}^{n}(x^{2})]](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B1%7D%7B2%7D%20%5B%5Csum_%7Bx%3D1%7D%5E%7Bn%7D%20%28x%29%20%2B%20%5Csum_%7Bx%3D1%7D%5E%7Bn%7D%28x%5E%7B2%7D%29%5D)
in this formula, for the first term, it's just an AP from x = 1 to x = n
for the second term, we have a general formula 
![\frac{1}{2}[\frac{n}{2}(2a + (n-1)d)+ \frac{n(n+1)(2n+1)}{6} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B%5Cfrac%7Bn%7D%7B2%7D%282a%20%2B%20%28n-1%29d%29%2B%20%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%7D%7B6%7D%20%20%5D)
in this AP (first term), the first term and the common difference is 1 as well
![\frac{1}{2}[\frac{n}{2}(2 + n-1)+ \frac{n(n+1)(2n+1)}{6} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B%5Cfrac%7Bn%7D%7B2%7D%282%20%2B%20n-1%29%2B%20%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%7D%7B6%7D%20%20%5D)
![\frac{1}{2}[\frac{n}{2}(n+1)+ \frac{n(n+1)(2n+1)}{6} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B%5Cfrac%7Bn%7D%7B2%7D%28n%2B1%29%2B%20%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%7D%7B6%7D%20%20%5D)
![[\frac{n}{4}(n+1)+ \frac{n(n+1)(2n+1)}{12} ]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bn%7D%7B4%7D%28n%2B1%29%2B%20%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%7D%7B12%7D%20%20%5D)
![\frac{n}{4}(n+1) [1+\frac{(2n+1)}{3} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bn%7D%7B4%7D%28n%2B1%29%20%5B1%2B%5Cfrac%7B%282n%2B1%29%7D%7B3%7D%20%5D)
![\frac{n}{4}(n+1) [\frac{(3+2n+1)}{3} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bn%7D%7B4%7D%28n%2B1%29%20%5B%5Cfrac%7B%283%2B2n%2B1%29%7D%7B3%7D%20%5D)
![\frac{n}{4}(n+1) [\frac{(2n+4)}{3} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bn%7D%7B4%7D%28n%2B1%29%20%5B%5Cfrac%7B%282n%2B4%29%7D%7B3%7D%20%5D)
![\frac{n}{2}(n+1) [\frac{(n+2)}{3} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bn%7D%7B2%7D%28n%2B1%29%20%5B%5Cfrac%7B%28n%2B2%29%7D%7B3%7D%20%5D)
which is the sum of n terms of the given sequence
It’s C. If you place the numbers from least to greatest, 126 (C) will be in the middle, or center.
Answer:
x =- 3 and y = -7
Step-by-step explanation:
2X-2Y=-8
X=2y + 11
We need to isolate both X and Y in both equations
so
2x-2y=-8
(add 2y to both sides)
2x=-8+ 2y
(divide both sides by 2)
x=-4+y and x=2y+11
because <u>both of these equations are the same </u>we can put them together
4+y=2y+11
(subtract y)
4=y+11
(subtract 11)
-7=y
so y = -7
then to find x you just need to plug in y to one of the equations
x=2(-7) + 11
x= -14 +11
x = -3
Answer:
x=4
Step-by-step explanation:
1+3x=2x+5
4=x
