There's a slight problem with your question, but we'll get to that...
Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the <em>n</em>-th term of the sequence, <em>a</em> (<em>n</em>), is given recursively by
• <em>a</em> (1) = 1
• <em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 21 … … … for <em>n</em> > 1
We can find the explicit rule for the sequence by iterative substitution:
<em>a</em> (2) = <em>a</em> (1) + 21
<em>a</em> (3) = <em>a</em> (2) + 21 = (<em>a</em> (1) + 21) + 21 = <em>a</em> (1) + 2×21
<em>a</em> (4) = <em>a</em> (3) + 21 = (<em>a</em> (1) + 2×21) + 21 = <em>a</em> (1) + 3×21
and so on, with the general pattern
<em>a</em> (<em>n</em>) = <em>a</em> (1) + 21 (<em>n</em> - 1) = 21<em>n</em> - 20
Now, we're told that the sum of some number <em>N</em> of terms in this sequence is 2332. In other words, the <em>N</em>-th partial sum of the sequence is
<em>a</em> (1) + <em>a</em> (2) + <em>a</em> (3) + … + <em>a</em> (<em>N</em> - 1) + <em>a</em> (<em>N</em>) = 2332
or more compactly,
![\displaystyle\sum_{n=1}^N a(n) = 2332](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5EN%20a%28n%29%20%3D%202332)
It's important to note that <em>N</em> must be some positive integer.
Replace <em>a</em> (<em>n</em>) by the explicit rule:
![\displaystyle\sum_{n=1}^N (21n-20) = 2332](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5EN%20%2821n-20%29%20%3D%202332)
Expand the sum on the left as
![\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332](https://tex.z-dn.net/?f=%5Cdisplaystyle%2021%20%5Csum_%7Bn%3D1%7D%5EN%20n-20%5Csum_%7Bn%3D1%7D%5EN1%20%3D%202332)
and recall the formulas,
![\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5En1%3D%5Cunderbrace%7B1%2B1%2B%5Ccdots%2B1%7D_%7Bn%5Ctext%7B%20times%7D%7D%3Dn)
![\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5Enk%3D1%2B2%2B3%2B%5Ccdots%2Bn%3D%5Cfrac%7Bn%28n%2B1%29%7D2)
So the sum of the first <em>N</em> terms of <em>a</em> (<em>n</em>) is such that
21 × <em>N</em> (<em>N</em> + 1)/2 - 20<em>N</em> = 2332
Solve for <em>N</em> :
21 (<em>N</em> ² + <em>N</em>) - 40<em>N</em> = 4664
21 <em>N</em> ² - 19 <em>N</em> - 4664 = 0
Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead
<em>N</em> = (19 ± √392,137)/42 ≈ -14.45 or 15.36
so there is no positive integer <em>N</em> for which the first <em>N</em> terms of the sum add up to 2332.