My teacher gave me a problem. Let \(1+4+10+20+35+..........+n=an^3-bn^2+cn-d\). If \(\frac{b}{a}+ \frac{d}{c}= \frac{e}{f}\) and \(gcd(e,f)=1\). What is the value of \(gcd(e+f,171)\)?

No vote yet

2 votes

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestMy answer is gcd(e + f, 171) = 1.

Solution: First, we must know that 1, 4, 10, 20, 35, ... form a kind of sequence. How do I know this? The sequence has a common difference in the third iteration or third operation of finding their common difference. (Sorry, I don't know how to type in Latex so I have to save time typing instead of typing and adjusting the texts). This sequence is ruled by the polynomial P(n) = (1/6) n^3 + (1/2) n^2 + (1/3) n.

Now to attack the question, since we have a sum of this sequence, therefore we have to sum up P(n). (Sorry again for cheating by using the formulas, I don't have much knowledge and mastery of Abel's Formula for Derivation.)

Since summation of n = (1/2)(n)(n+1) Summation of n^2 = (1/6)(n)(n+1)(2n+1) Summation of n^3 = ((1/2)(n)(n+1))^2

To simplify this up, Summation of P(n) = (1/24)(n^4 + 6 n^3 + 11 n^2 + 6 n) Equating this to an^3-bn^2+cn-d, we have to make equivalences since the summation of P(n) has degree of 4 while the given has 3.

This will yield: a = n/24 b = -n/4 c = 11n/24 d = -n/4 And checking these values will be equivalent to P(n).

Now to evaluate b/a+d/c=e/f (-n/4)(24/n) + (-n/4)(24/11n) = e/f e/f = -72/11 gcd (-72, 11) = 1 So e + f = -61 or 61 (This depends on where you will assign the negative sign but it does not have an effect on the gcd of these numbers.)

Final Answer: gcd(-61, 171) = gcd(61, 171) = 1 – John Ashley Capellan · 3 years, 9 months ago

Log in to reply

Log in to reply