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 n-th term of the sequence, a (n), is given recursively by
• a (1) = 1
• a (n) = a (n - 1) + 21 … … … for n > 1
We can find the explicit rule for the sequence by iterative substitution:
a (2) = a (1) + 21
a (3) = a (2) + 21 = (a (1) + 21) + 21 = a (1) + 2×21
a (4) = a (3) + 21 = (a (1) + 2×21) + 21 = a (1) + 3×21
and so on, with the general pattern
a (n) = a (1) + 21 (n - 1) = 21n - 20
Now, we're told that the sum of some number N of terms in this sequence is 2332. In other words, the N-th partial sum of the sequence is
a (1) + a (2) + a (3) + … + a (N - 1) + a (N) = 2332
or more compactly,
[tex]\displaystyle\sum_{n=1}^N a(n) = 2332[/tex]
It's important to note that N must be some positive integer.
Replace a (n) by the explicit rule:
[tex]\displaystyle\sum_{n=1}^N (21n-20) = 2332[/tex]
Expand the sum on the left as
[tex]\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332[/tex]
and recall the formulas,
[tex]\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n[/tex]
[tex]\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2[/tex]
So the sum of the first N terms of a (n) is such that
21 × N (N + 1)/2 - 20N = 2332
Solve for N :
21 (N ² + N) - 40N = 4664
21 N ² - 19 N - 4664 = 0
Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead
N = (19 ± √392,137)/42 ≈ -14.45 or 15.36
so there is no positive integer N for which the first N terms of the sum add up to 2332.
