
Steve S. answered 03/31/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
f(n) = (4n+3)!/(4n-3)!
= ((4n+3)(4n+2)(4n+1)(4n)(4n-1)(4n-2)) (4n-3)!/(4n-3)!
f(n) = (4n+3)(4n+2)(4n+1)(4n)(4n-1)(4n-2)
====
Optional: I’m not sure why you would want to do this, but a professor once nonchalantly said to me in a class that to multiply the factors you could use the distributive property to multiply all combinations of the binomial terms. So I thought I’d try it here (retired, you know –– hey, it’s better than Sudoku! ;-) ).
Since you can copy lines easily, it’s really not too hard. A little tedious. But I got the right answer!
f(n) =
+ (4n)(4n)(4n)(4n)(4n)(4n) | (4n)^6
+ (4n)(4n)(4n)(4n)(4n)(-2) | –2(4n)^5
+ (4n)(4n)(4n)(4n)(-1)(4n) | –(4n)^5
+ (4n)(4n)(4n)(4n)(-1)(-2) | 2(4n)^4
+ (4n)(4n)(4n)(+0)(4n)(4n) | 0
+ (4n)(4n)(4n)(+0)(4n)(-2) | 0
+ (4n)(4n)(4n)(+0)(-1)(4n) | 0
+ (4n)(4n)(4n)(+0)(-1)(-2) | 0
+ (4n)(4n)(+1)(4n)(4n)(4n) | (4n)^5
+ (4n)(4n)(+1)(4n)(4n)(-2) | –2(4n)^4
+ (4n)(4n)(+1)(4n)(-1)(4n) | –(4n)^4
+ (4n)(4n)(+1)(4n)(-1)(-2) | 2(4n)^3
+ (4n)(4n)(+1)(+0)(4n)(4n) | 0
+ (4n)(4n)(+1)(+0)(4n)(-2) | 0
+ (4n)(4n)(+1)(+0)(-1)(4n) | 0
+ (4n)(4n)(+1)(+0)(-1)(-2) | 0
+ (4n)(+2)(4n)(4n)(4n)(4n) | 2(4n)^5
+ (4n)(+2)(4n)(4n)(4n)(-2) | –4(4n)^4
+ (4n)(+2)(4n)(4n)(-1)(4n) | –2(4n)^4
+ (4n)(+2)(4n)(4n)(-1)(-2) | 4(4n)^3
+ (4n)(+2)(4n)(+0)(4n)(4n) | 0
+ (4n)(+2)(4n)(+0)(4n)(-2) | 0
+ (4n)(+2)(4n)(+0)(-1)(4n) | 0
+ (4n)(+2)(4n)(+0)(-1)(-2) | 0
+ (4n)(+2)(+1)(4n)(4n)(4n) | 2(4n)^4
+ (4n)(+2)(+1)(4n)(4n)(-2) | –4(4n)^3
+ (4n)(+2)(+1)(4n)(-1)(4n) | –2(4n)^3
+ (4n)(+2)(+1)(4n)(-1)(-2) | 4(4n)^2
+ (4n)(+2)(+1)(+0)(4n)(4n) | 0
+ (4n)(+2)(+1)(+0)(4n)(-2) | 0
+ (4n)(+2)(+1)(+0)(-1)(4n) | 0
+ (4n)(+2)(+1)(+0)(-1)(-2) | 0
+ (+3)(4n)(4n)(4n)(4n)(4n) | 3(4n)^5
+ (+3)(4n)(4n)(4n)(4n)(-2) | –6(4n)^4
+ (+3)(4n)(4n)(4n)(-1)(4n) | –3(4n)^4
+ (+3)(4n)(4n)(4n)(-1)(-2) | 6(4n)^3
+ (+3)(4n)(4n)(+0)(4n)(4n) | 0
+ (+3)(4n)(4n)(+0)(4n)(-2) | 0
+ (+3)(4n)(4n)(+0)(-1)(4n) | 0
+ (+3)(4n)(4n)(+0)(-1)(-2) | 0
+ (+3)(4n)(+1)(4n)(4n)(4n) | 3(4n)^4
+ (+3)(4n)(+1)(4n)(4n)(-2) | –6(4n)^3
+ (+3)(4n)(+1)(4n)(-1)(4n) | –3(4n)^3
+ (+3)(4n)(+1)(4n)(-1)(-2) | 6(4n)^2
+ (+3)(4n)(+1)(+0)(4n)(4n) | 0
+ (+3)(4n)(+1)(+0)(4n)(-2) | 0
+ (+3)(4n)(+1)(+0)(-1)(4n) | 0
+ (+3)(4n)(+1)(+0)(-1)(-2) | 0
+ (+3)(+2)(4n)(4n)(4n)(4n) | 6(4n)^4
+ (+3)(+2)(4n)(4n)(4n)(-2) | –12(4n)^3
+ (+3)(+2)(4n)(4n)(-1)(4n) | –6(4n)^3
+ (+3)(+2)(4n)(4n)(-1)(-2) | 12(4n)^2
+ (+3)(+2)(4n)(+0)(4n)(4n) | 0
+ (+3)(+2)(4n)(+0)(4n)(-2) | 0
+ (+3)(+2)(4n)(+0)(-1)(4n) | 0
+ (+3)(+2)(4n)(+0)(-1)(-2) | 0
+ (+3)(+2)(+1)(4n)(4n)(4n) | 6(4n)^3
+ (+3)(+2)(+1)(4n)(4n)(-2) | –12(4n)^2
+ (+3)(+2)(+1)(4n)(-1)(4n) | –6(4n)^2
+ (+3)(+2)(+1)(4n)(-1)(-2) | 12(4n)
+ (+3)(+2)(+1)(+0)(4n)(4n) | 0
+ (+3)(+2)(+1)(+0)(4n)(-2) | 0
+ (+3)(+2)(+1)(+0)(-1)(4n) | 0
+ (+3)(+2)(+1)(+0)(-1)(-2) | 0
f(n) = (4n)^6+3(4n)^5–5(4n)^4–15(4n)^3+4(4n)^2+12(4n)
f(n) = 2^12 n^6 +3 2^10 n^5 –5 2^8 n^4 –15 2^6 n^3 +2^6 n^2 +48 n
f(n) =
4096 n^6 +3072 n^5 –1280 n^4 –960 n^3 +64 n^2 +48 n
= (4n+3)(4n+2)(4n+1)(4n)(4n-1)(4n-2) √
Checked with GeoGebra.
= ((4n+3)(4n+2)(4n+1)(4n)(4n-1)(4n-2)) (4n-3)!/(4n-3)!
f(n) = (4n+3)(4n+2)(4n+1)(4n)(4n-1)(4n-2)
====
Optional: I’m not sure why you would want to do this, but a professor once nonchalantly said to me in a class that to multiply the factors you could use the distributive property to multiply all combinations of the binomial terms. So I thought I’d try it here (retired, you know –– hey, it’s better than Sudoku! ;-) ).
Since you can copy lines easily, it’s really not too hard. A little tedious. But I got the right answer!
f(n) =
+ (4n)(4n)(4n)(4n)(4n)(4n) | (4n)^6
+ (4n)(4n)(4n)(4n)(4n)(-2) | –2(4n)^5
+ (4n)(4n)(4n)(4n)(-1)(4n) | –(4n)^5
+ (4n)(4n)(4n)(4n)(-1)(-2) | 2(4n)^4
+ (4n)(4n)(4n)(+0)(4n)(4n) | 0
+ (4n)(4n)(4n)(+0)(4n)(-2) | 0
+ (4n)(4n)(4n)(+0)(-1)(4n) | 0
+ (4n)(4n)(4n)(+0)(-1)(-2) | 0
+ (4n)(4n)(+1)(4n)(4n)(4n) | (4n)^5
+ (4n)(4n)(+1)(4n)(4n)(-2) | –2(4n)^4
+ (4n)(4n)(+1)(4n)(-1)(4n) | –(4n)^4
+ (4n)(4n)(+1)(4n)(-1)(-2) | 2(4n)^3
+ (4n)(4n)(+1)(+0)(4n)(4n) | 0
+ (4n)(4n)(+1)(+0)(4n)(-2) | 0
+ (4n)(4n)(+1)(+0)(-1)(4n) | 0
+ (4n)(4n)(+1)(+0)(-1)(-2) | 0
+ (4n)(+2)(4n)(4n)(4n)(4n) | 2(4n)^5
+ (4n)(+2)(4n)(4n)(4n)(-2) | –4(4n)^4
+ (4n)(+2)(4n)(4n)(-1)(4n) | –2(4n)^4
+ (4n)(+2)(4n)(4n)(-1)(-2) | 4(4n)^3
+ (4n)(+2)(4n)(+0)(4n)(4n) | 0
+ (4n)(+2)(4n)(+0)(4n)(-2) | 0
+ (4n)(+2)(4n)(+0)(-1)(4n) | 0
+ (4n)(+2)(4n)(+0)(-1)(-2) | 0
+ (4n)(+2)(+1)(4n)(4n)(4n) | 2(4n)^4
+ (4n)(+2)(+1)(4n)(4n)(-2) | –4(4n)^3
+ (4n)(+2)(+1)(4n)(-1)(4n) | –2(4n)^3
+ (4n)(+2)(+1)(4n)(-1)(-2) | 4(4n)^2
+ (4n)(+2)(+1)(+0)(4n)(4n) | 0
+ (4n)(+2)(+1)(+0)(4n)(-2) | 0
+ (4n)(+2)(+1)(+0)(-1)(4n) | 0
+ (4n)(+2)(+1)(+0)(-1)(-2) | 0
+ (+3)(4n)(4n)(4n)(4n)(4n) | 3(4n)^5
+ (+3)(4n)(4n)(4n)(4n)(-2) | –6(4n)^4
+ (+3)(4n)(4n)(4n)(-1)(4n) | –3(4n)^4
+ (+3)(4n)(4n)(4n)(-1)(-2) | 6(4n)^3
+ (+3)(4n)(4n)(+0)(4n)(4n) | 0
+ (+3)(4n)(4n)(+0)(4n)(-2) | 0
+ (+3)(4n)(4n)(+0)(-1)(4n) | 0
+ (+3)(4n)(4n)(+0)(-1)(-2) | 0
+ (+3)(4n)(+1)(4n)(4n)(4n) | 3(4n)^4
+ (+3)(4n)(+1)(4n)(4n)(-2) | –6(4n)^3
+ (+3)(4n)(+1)(4n)(-1)(4n) | –3(4n)^3
+ (+3)(4n)(+1)(4n)(-1)(-2) | 6(4n)^2
+ (+3)(4n)(+1)(+0)(4n)(4n) | 0
+ (+3)(4n)(+1)(+0)(4n)(-2) | 0
+ (+3)(4n)(+1)(+0)(-1)(4n) | 0
+ (+3)(4n)(+1)(+0)(-1)(-2) | 0
+ (+3)(+2)(4n)(4n)(4n)(4n) | 6(4n)^4
+ (+3)(+2)(4n)(4n)(4n)(-2) | –12(4n)^3
+ (+3)(+2)(4n)(4n)(-1)(4n) | –6(4n)^3
+ (+3)(+2)(4n)(4n)(-1)(-2) | 12(4n)^2
+ (+3)(+2)(4n)(+0)(4n)(4n) | 0
+ (+3)(+2)(4n)(+0)(4n)(-2) | 0
+ (+3)(+2)(4n)(+0)(-1)(4n) | 0
+ (+3)(+2)(4n)(+0)(-1)(-2) | 0
+ (+3)(+2)(+1)(4n)(4n)(4n) | 6(4n)^3
+ (+3)(+2)(+1)(4n)(4n)(-2) | –12(4n)^2
+ (+3)(+2)(+1)(4n)(-1)(4n) | –6(4n)^2
+ (+3)(+2)(+1)(4n)(-1)(-2) | 12(4n)
+ (+3)(+2)(+1)(+0)(4n)(4n) | 0
+ (+3)(+2)(+1)(+0)(4n)(-2) | 0
+ (+3)(+2)(+1)(+0)(-1)(4n) | 0
+ (+3)(+2)(+1)(+0)(-1)(-2) | 0
f(n) = (4n)^6+3(4n)^5–5(4n)^4–15(4n)^3+4(4n)^2+12(4n)
f(n) = 2^12 n^6 +3 2^10 n^5 –5 2^8 n^4 –15 2^6 n^3 +2^6 n^2 +48 n
f(n) =
4096 n^6 +3072 n^5 –1280 n^4 –960 n^3 +64 n^2 +48 n
= (4n+3)(4n+2)(4n+1)(4n)(4n-1)(4n-2) √
Checked with GeoGebra.
Dalia S.
04/01/14