One day Shahed was playing with numbers. He wrote 15 fractions using all natural numbers from 1 to 30 exactly once- either as nu
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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,
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:
Expand the sum on the left as
and recall the formulas,
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.