SOLUTION: Find the arithmetic sequence which has the sum of its n terms equal to 2n^2+3n Thank you

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Question 1200515: Find the arithmetic sequence which has the sum of its n terms equal to 2n^2+3n
Thank you
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
Find the arithmetic sequence which has the sum of its n terms equal to 2n^2+3n
~~~~~~~~~~~~~~~~~~~~~~~ 

To solve this problem, you need to know that 


    +----------------------------------------------------+
    |    the sum of any arithmetic progression is        |
    |      a quadratic function of the index "n",        | 
    +----------------------------------------------------+


and vice versa,


    +----------------------------------------------------+
    |   any quadratic function of the integer index "n"  |
    |      creates a unique arithmetic progression.      |
    +----------------------------------------------------+


Based on these facts, it is enough to find the first term of the progression
and the second term, and then calculate its common difference.


    n= 1:   = 2*1^2+ 3*1 = 5.      = 5.


    n= 2:   = 2*2^2+ 3*2 = 14.     =  -  = 14 - 5 = 9.


This arithmetic progression has the first term   = 5  and the common difference d=  -  = 9 - 5 = 4.


The progression is  5, 9, 13, 17, . . .

Solved.

-------------------

On these properties of arithmetic progression, learn from the lessons
    - Free fall and arithmetic progressions
    - Uniformly accelerated motions and arithmetic progressions
    - Increments of a quadratic function form an arithmetic progression
in this site.



Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Answer: 5, 9, 13, 17, 21, ...
Starting term = 5
Common difference = 4
The nth term formula is a(n) = 5+4*(n-1) which fully simplifies to a(n) = 4n+1


Explanation:

Sn = sum of the first n terms
Sn = 2n^2 + 3n
This was given to us in the instructions.

If n = 1, then Sn = S1 = the first term of the arithmetic sequence.
Let's find that first term.
Sn = 2n^2 + 3n
S1 = 2*1^2 + 3*1
S1 = 2*1 + 3*1
S1 = 2 + 3
S1 = 5
The first term of the arithmetic sequence is 5.

Now let's find S2
Sn = 2n^2 + 3n
S2 = 2*2^2 + 3*2
S2 = 2*4 + 3*2
S2 = 8 + 6
S2 = 14
This represents the sum of the first two terms. We'll use it to determine what the second term would be:
S2 = term1+term2
14 = 5+term2
term2 = 14-5
term2 = 9

The arithmetic sequence so far is 5,9,...
This is sufficient info to determine the common difference
d = term2-term1
d = 9-5
d = 4

So we have:
a1 = first term = 5
d = common difference = 4
The nth term formula is:
a(n) = a1 + d*(n-1)
a(n) = 5 + 4*(n-1)
a(n) = 5 + 4n-4
a(n) = 4n + 1


Check:
n (method 1) (method 2)
1552n^2+3n = 2*1^2+3*1 = 5
295+9 = 142n^2+3n = 2*2^2+3*2 = 14
3135+9+13 = 272n^2+3n = 2*3^2+3*3 = 27
4175+9+13+17 = 442n^2+3n = 2*4^2+3*4 = 44
5215+9+13+17+21 = 652n^2+3n = 2*5^2+3*5 = 65

The two columns have their results match up, so this helps confirm the answer.

Another way to check:
One formula to quickly find the sum of the first n terms of an arithmetic sequence

Plug in the first term of 5

Plug in the nth term formula of a(n) = 4n+1





Therefore, the answer is confirmed.
Arguably this is better confirmation compared to the table of examples as shown above

Another possible arithmetic summation formula to use would be .
Plug in and . Simplifying should lead to
I'll let you do this verification pathway.
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