Question textConsider the following sequence for which three terms are shown:p=3!1!+5!3!+7!5!+…p=\frac{3!}{1!}+\frac{5!}{3!}+\frac{7!}{5!}+\dotsIn general: pi=(2i+1)!(2i−1)!=(2i)(2i+1)p_i=\frac{(2i+1)!}{(2i-1)!}=(2i)(2i+1)What is the minimum number of terms required for p to be greater than 131,313? Answer 1 Question 2[input]

Question

BlackTom AI · Accepted Answer

Rewriting the problem in my own words: We have a sum p consisting of terms of the form (2i+1)!/(2i-1)!, which simplifies to (2i)(2i+1). The question asks for the minimum number of terms n needed so that the partial sum p > 131,313.

Step 1: Express each term in a simpler form.
- (2i+1)!/(2i-1)! = (2i)(2i+1).
Thus the nth term is T_i = (2i)(2i+1) = 4i^2 + 2i.

Step 2: Sum the first n terms.
- S(n) = Σ_{i=1 to n} (4i^2 + 2i) = 4 Σ i^2 + 2 Σ i.
We use the standard formulas: Σ i = n(n+1)/2 and Σ i^2 = n(n+1)(2n+1)/6.
So S(n) = 4 * [n(n+1)(2n+1)/6] + 2 * [n(n+1)/2] = [2/3] n(n+1)(2n+1) + n(n+1).
Factor out n(n+1): S(n) = n(n+1) [ (2n+1)/? wait, compute carefully: (2/3)(2n+1) + 1 = (4n+2)/3 + 1 = (4n+5)/3. Therefore, S(n) = n(n+1)(4n+5)/3.

Step 3: Find the smallest n such that S(n) > 131,313.
- We need n(n+1)(4n+5) > 3 * 131,313 = 393,939.
Test n = 45:
  S(45) = 45*46*(4*45+5)/3 = 45*46*185/3 = (45*46/3)*185 = 690*185 = 127,650, which is below 131,313.
Test n = 46:
  S(46) = 46*47*(4*46+5)/3 = 46*47*189/3 = (46*47/3)*189 = 1,? Compute step by step: 46*47 = 2,162; 2,162*189 = 2,162*(200-11) = 432,400 - 23,782 = 408,618; divide by 3 gives 136,206. This exceeds 131,313.
Since S(45) is too small and S(46) is large enough, the transition occurs at n = 46.

Step 4: Conclusion drawn from the comparisons above.
- Before reaching 46 terms, the sum is still less than 131,313.
- With 46 terms, the sum surpasses 131,313.

Therefore, the minimum number of terms required for p to exceed 131,313 is reached at 46 terms, as demonstrated by the computed partial sums.

类似问题

Question textConsider the following sequence for which three terms are shown:[math: p=3!1!+5!3!+7!5!+…]In general: [math: pi=(2i+1)!(2i−1)!=(2i)(2i+1)]What is the minimum number of terms required for p to be greater than 131,313? Answer 1 Question 2[input]Check Question 2

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