Check the forward differences of the sequence.
If
, then let
be the sequence of first-order differences of
. That is, for n ≥ 1,
![b_n = a_{n+1} - a_n](https://tex.z-dn.net/?f=b_n%20%3D%20a_%7Bn%2B1%7D%20-%20a_n)
so that
.
Let
be the sequence of differences of
,
![c_n = b_{n+1} - b_n](https://tex.z-dn.net/?f=c_n%20%3D%20b_%7Bn%2B1%7D%20-%20b_n)
and we see that this is a constant sequence,
. In other words,
is an arithmetic sequence with common difference between terms of 2. That is,
![2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2](https://tex.z-dn.net/?f=2%20%3D%20b_%7Bn%2B1%7D%20-%20b_n%20%5Cimplies%20b_%7Bn%2B1%7D%20%3D%20b_n%20%2B%202)
and we can solve for
in terms of
:
![b_{n+1} = b_n + 2](https://tex.z-dn.net/?f=b_%7Bn%2B1%7D%20%3D%20b_n%20%2B%202)
![b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2](https://tex.z-dn.net/?f=b_%7Bn%2B1%7D%20%3D%20%28b_%7Bn-1%7D%2B2%29%20%2B%202%20%3D%20b_%7Bn-1%7D%20%2B%202%5Ctimes2)
![b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2](https://tex.z-dn.net/?f=b_%7Bn%2B1%7D%20%3D%20%28b_%7Bn-2%7D%2B2%29%20%2B%202%5Ctimes2%20%3D%20b_%7Bn-2%7D%20%2B%203%5Ctimes2)
and so on down to
![b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)](https://tex.z-dn.net/?f=b_%7Bn%2B1%7D%20%3D%20b_1%20%2B%202n%20%5Cimplies%20b_%7Bn%2B1%7D%20%3D%202n%20%2B%204%20%5Cimplies%20b_n%20%3D%202%28n-1%29%2B4%20%3D%202%28n%20%2B%201%29)
We solve for
in the same way.
![2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)](https://tex.z-dn.net/?f=2%28n%2B1%29%20%3D%20a_%7Bn%2B1%7D%20-%20a_n%20%5Cimplies%20a_%7Bn%2B1%7D%20%3D%20a_n%20%2B%202%28n%20%2B%201%29)
Then
![a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)](https://tex.z-dn.net/?f=a_%7Bn%2B1%7D%20%3D%20%28a_%7Bn-1%7D%20%2B%202n%29%20%2B%202%28n%2B1%29%20%5C%5C%20~~~~~~~%3D%20a_%7Bn-1%7D%20%2B%202%20%28%28n%2B1%29%20%2B%20n%29)
![a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))](https://tex.z-dn.net/?f=a_%7Bn%2B1%7D%20%3D%20%28a_%7Bn-2%7D%20%2B%202%28n-1%29%29%20%2B%202%28%28n%2B1%29%2Bn%29%20%5C%5C%20~~~~~~~%20%3D%20a_%7Bn-2%7D%20%2B%202%20%28%28n%2B1%29%20%2B%20n%20%2B%20%28n-1%29%29)
![a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))](https://tex.z-dn.net/?f=a_%7Bn%2B1%7D%20%3D%20%28a_%7Bn-3%7D%20%2B%202%28n-2%29%29%20%2B%202%28%28n%2B1%29%2Bn%2B%28n-1%29%29%20%5C%5C%20~~~~~~~%3D%20a_%7Bn-3%7D%20%2B%202%20%28%28n%2B1%29%20%2B%20n%20%2B%20%28n-1%29%20%2B%20%28n-2%29%29)
and so on down to
![a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2](https://tex.z-dn.net/?f=a_%7Bn%2B1%7D%20%3D%20a_1%20%2B%202%20%5Cdisplaystyle%20%5Csum_%7Bk%3D2%7D%5E%7Bn%2B1%7D%20k%20%3D%202%20%2B%202%20%5Ctimes%20%5Cfrac%7Bn%28n%2B3%29%7D2)
![\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}](https://tex.z-dn.net/?f=%5Cimplies%20a_%7Bn%2B1%7D%20%3D%20n%5E2%20%2B%203n%20%2B%202%20%5Cimplies%20%5Cboxed%7Ba_n%20%3D%20n%5E2%20%2B%20n%7D)
its the first one
Step-by-step explanation:
because per pertains to to x
Answer: 530.66 units²
<u>Step-by-step explanation:</u>
![Area_{(circle)}=\pi r^2\qquad and\qquad r=\dfrac{diameter}{2}\\\\\\\implies Area_{(circle)}=\pi \bigg(\dfrac{diameter}{2}\bigg)^2\\\\\\\text{It is given that the diameter = 26:}\\A=\pi\bigg(\dfrac{26}{2}\bigg)^2\\\\\\.\quad =\pi \cdot 13^2\\\\\\.\quad =169\pi\\\\\\.\quad =\large\boxed{530.66}](https://tex.z-dn.net/?f=Area_%7B%28circle%29%7D%3D%5Cpi%20r%5E2%5Cqquad%20and%5Cqquad%20r%3D%5Cdfrac%7Bdiameter%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5Cimplies%20Area_%7B%28circle%29%7D%3D%5Cpi%20%5Cbigg%28%5Cdfrac%7Bdiameter%7D%7B2%7D%5Cbigg%29%5E2%5C%5C%5C%5C%5C%5C%5Ctext%7BIt%20is%20given%20that%20the%20diameter%20%3D%2026%3A%7D%5C%5CA%3D%5Cpi%5Cbigg%28%5Cdfrac%7B26%7D%7B2%7D%5Cbigg%29%5E2%5C%5C%5C%5C%5C%5C.%5Cquad%20%3D%5Cpi%20%5Ccdot%2013%5E2%5C%5C%5C%5C%5C%5C.%5Cquad%20%3D169%5Cpi%5C%5C%5C%5C%5C%5C.%5Cquad%20%3D%5Clarge%5Cboxed%7B530.66%7D)
7/8 is ur answer because
it can be cut into smaller parts