Sum of the first n terms of the series whose nth term is 3n^2 + n + 1

Here is the solution for the sum $$S_1 = n(3n^2) = 3n^3$$ Let $S_2 = n$ + sum of $5n, 11n, 17n, 23n \ldots$. Take note that the number of terms of $5n, 11n, 17n, 23n \ldots$ is $n - 1$ only, not $n$. Note that the sum of AP is given by $S = \frac{n}{2} \left[ 2a_1 + (n - 1)d \right]$. Thus $$S_2 = n + \frac{n - 1}{2} \{ 2(5n) + [(n - 1) - 1]6 \}$$ $$S_2 = n + \frac{n - 1}{2} \left[ 10n + (6n - 12) \right]$$ $$S_2 = n + \frac{n - 1}{2} (16n - 12)$$ $$S_2 = n + (n - 1)(8n - 6)$$ $$S_2 = n + 2(n - 1)(4n - 3)$$ The sum of constants $$S_3 = 1 + 3 + 11 + 25 + 45 + \ldots$$ Thus the sum of the first $n$ terms of $3n^2 + n + 1$ is ... $$S = S_1 - S_2 + S_3$$ $$S = 3n^3 - [n + 2(n - 1)(4n - 3)] + (1 + 3 + 11 + 25 + 45 + \ldots )$$ I hope somebody can share to us the generalized value of $S_3$. I'm also looking forward on you Nacheez.

Hi Nickson,

Thank you for your post, it made me reconsider my solution in Reply #1 of this thread. For the sake of clarity, I will rewrite my solution from Reply #1.

Linje 1: $a_n = 3n^2 + n + 1$
Linje 2: $a_{n-1} = 3(n - 1)^2 + (n - 1) + 1 = 3n^2 - 5n + 3$
Linje 3: $a_{n-2} = 3(n - 2)^2 + (n - 2) + 1 = 3n^2 - 11n + 11$
Linje 4: $a_{n-3} = 3(n - 3)^2 + (n - 3) + 1 = 3n^2 - 17n + 25$
Linje 5: $a_{n-4} = 3(n - 4)^2 + (n - 4) + 1 = 3n^2 - 23n + 45$
Linje 6: $\vdots$
Linje 7: $a_4 = 3(4^2) + 4 + 1 = 53$
Linje 8: $a_3 = 31$
Linje 9: $a_2 = 15$
Linje 10: $a_1 = 5$

The problem with the above solution is the merging of quantities of first and second terms that both involves n. You can take a look at lines 2 to 5 wehrein the quantities involving n were merged. Example is the line 2: $$a_{n-1} = 3(n - 1)^2 + (n - 1) + 1 = 3n^2 - 5n + 3$$ where the "n" of (n - 1)² was merged to "n" of (n - 1). This makes the generalization of the final answer more complex and hard to do. With the help of your post, I made a solution that sums up the first terms, sum up the second terms, sum up the third terms, then take the summation of the sum of the first, second, and third terms. This is exactly how you solved the problem, and in fact I arrived with the formula you use.

Here is my new solution:
$a_n = 3n^2 + n + 1$
$a_{n-1} = 3(n - 1)^2 + (n - 1) + 1$
$a_{n-2} = 3(n - 2)^2 + (n - 2) + 1$
$a_{n-3} = 3(n - 3)^2 + (n - 3) + 1$
$\vdots$

And now for the main event; finding the sum of all these terms.
$S = \Big[ 3n^2 + 3(n - 1)^2 +... + 3(2^2) + 3(1^2) \Big] + \Big[ n + (n - 1) +... + 2 + 1 \Big] + n(1)$
$S = 3 \Big[ n^2 + (n - 1)^2 +... + (2^2) + (1^2) \Big] + \Big[ n + (n - 1) +... + 2 + 1 \Big] + n$
$S = 3 \Big[ 1^2 + 2^2 +... + (n - 1)^2 + n^2 \Big] + \Big[ 1 + 2 +... + (n - 1) + n \Big] + n$
$S = \sum_{n=1}^n n^2 + \sum_{n=1}^n n + n$

The $\sum_{n=1}^n n = 1 + 2 + 3 +... + (n - 2) + (n - 1) + n$ is an arithmetic progression with a common diffference of 1. Thus, $$\sum_{n=1}^n n = \frac{n}{2} (1 + n).$$ In case you are not familiar with the formula, it can easily be done by twice the sum as follows: $$\underline{\matrix{1&+&2&+&3&+&...&+&n-2&n-1&n\\n&+&n-1&+&n-2&+&...&+&3&2&1}}$$ $$\matrix{(1+n)&+&(1+n)&+&...&+&(1+n)&+&(1+n)}$$ Then twice the sum, $2S = (1+n)+(1+n)+ ...$ up to n count. Thus, $$2S = n(n + 1) \text{ OR}$$ $$S = \frac{n}{2} (1 + n) \text{ for } \sum_{n=1}^n n$$

For $\sum_{n=1}^n n^2$, we can generalize it by the following:
Let

I got to go, I will edit this tomorrow(?)

I decided not to continue the solution in the above reply but create this new reply to make it compact.

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Continuation...

For $\sum_{n=1}^n n^2$, we can generalize it by the following:
Consider:
$f(n) = (n + 1)^3 - n^3$
$f(n) = (n^3 + 3n^2 + 3n + 1) - n^3$
$f(n) = 3n^2 + 3n + 1$
We can take the summation of $f(n)$ from n = 1 to n = n in two ways:
FIRST:
$\sum_{n=1}^n f(n) = \sum_{n=1}^3 \Big[ (n + 1)^3 - n^3 \Big]$
$\sum_{n=1}^n f(n) = (2^3 - 1^3) + (3^3 - 2^3) +... + \Big[ n^3 - (n - 1)^3 \Big] + \Big[ (n + 1)^3 + n^3 \Big]$
By method of differences, we can see that...
$\sum_{n=1}^n f(n) = -1^3 + (n + 1)^3$
$\sum_{n=1}^n f(n) = -1^3 + (n^3 + 3n^2 + 3n + 1^3)$
$\sum_{n=1}^n f(n) = n^3 + 3n^2 + 3n$
SECOND:
$\sum_{n=1}^n f(n) = \sum_{n=1}^n (3n^2 + 3n + 1)$
$\sum_{n=1}^n f(n) = 3\sum_{n=1}^n n^2 + 3\sum_{n=1}^n n + \sum_{n=1}^n (1)$
Note that the value of $\sum_{n=1}^n n$ is shown above, thus...
$\sum_{n=1}^n f(n) = 3\sum_{n=1}^n n^2 + 3(n/2)(1 + n) + n$
$\sum_{n=1}^n f(n) = 3\sum_{n=1}^n n^2 + \frac{3}{2}n^2 + \frac{3}{2}n + n$
$\sum_{n=1}^n f(n) = 3\sum_{n=1}^n n^2 + \frac{3}{2}n^2 + \frac{5}{2}n$
Equating the values of $\sum_{n=1}^n f(n)$:
$3\sum_{n=1}^n n^2 + \frac{3}{2}n^2 + \frac{5}{2}n = n^3 + 3n^2 + 3n$
$3\sum_{n=1}^n n^2 = n^3 + \frac{3}{2}n^2 + \frac{1}{2}n$
$3\sum_{n=1}^n n^2 = \frac{1}{2} (2n^3 + 3n^2 + 1)$
$\sum_{n=1}^n n^2 = \frac{1}{6} (2n^3 + 3n^2 + 1)$ $$\sum_{n=1}^n n^2 = \frac{n(2n + 1)(n + 1)}{6}$$

Thus, the required sum is $$S = \sum_{n=1}^n n^2 + \sum_{n=1}^n n + n$$ as solved by Nickson. Thanks Nickson, your reply is great.

Cheers
RTFVerterra

Hi Sir Morpheus of EMC,
I like the way how you attack the problem; it is clean, short, and fast. I completely understand the how's but I have no idea about the why's. This is my first view of this kind of solution. Maybe you can give us some suggested readings for us to understand the principles underlying your solution. You may mention a topic or a subject, much better if you can give us some links. Thanks, I highly recommend your solution. I tried it for other equations, it works great. Cheers...